Edexcel A Level Maths: Mechanics

Topic Questions

2.4 Variable Acceleration - 2D (A Level only)

1a
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3 marks

A particle’s position, at time t seconds, is given by the vector

 

bold italic r space equals space open parentheses table row cell 2 t cubed space minus space 1 end cell row cell t squared space plus space 4 end cell end table close parentheses space m 

 

(i)
Find the coordinates of the starting position of the particle.

(ii)
Find the position vector of the particle after 6 seconds.
1b
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1 mark

Consider the y-coordinate of the particle to explain why the particle will never pass through the origin.

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2a
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2 marks

At time t seconds, a particle moving in a plane has velocity

 

bold italic v space equals space open parentheses open parentheses 2 t cubed space minus space 4 t close parentheses bold i space plus space open parentheses 2 t space minus 3 close parentheses bold j close parentheses space straight m space straight s to the power of negative 1 end exponent

 

Use differentiation to find an expression for the acceleration of the particle.

2b
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3 marks

Use integration to find the displacement of the particle from its initial position.

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3a
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2 marks

The acceleration of a particle is modelled using

 

bold italic a space equals space open parentheses open parentheses 2 space minus 8 t close parentheses bold i space plus space open parentheses 6 t squared close parentheses bold j close parentheses space straight m space straight s to the power of negative 2 end exponent

 

where time t is measured in seconds.

Given that the particle is initially at rest, use integration to find an expression for the velocity of the particle.

 

3b
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3 marks
(i)
Find the velocity of the particle at time t = 2 seconds

(ii)
Use Pythagoras’ theorem to show that the speed of the particle at time t equals 2 space seconds is 20 m s-1.

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4a
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2 marks

The position vector of a particle, at time t seconds, is given by

bold r space equals space open parentheses open parentheses sin space t close parentheses bold i space plus open parentheses cos space 2 t close parentheses bold j close parentheses space straight m space space space space space space space space space space space space space space space 0 space less or equal than space t space less or equal than space straight pi

Differentiate r with respect to t to find an expression for the velocity of the particle.

4b
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3 marks
(i)
Find the time at which the velocity of the particle in the i-direction is 0.5 space straight m space straight s to the power of negative 1 end exponent.

(ii)
Hence find the velocity in the j-direction at this time.

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5a
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2 marks

The velocity of a particle at time t seconds is given by

 

bold italic v space equals space open parentheses table row cell e to the power of t space minus space t end cell row cell 0.5 t to the power of 4 end cell end table close parentheses space straight m space straight s to the power of negative 1 end exponent

 

Given that the particle’s motion began at the origin, use integration to find the position vector of the particle at time t seconds.

5b
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2 marks

Use Pythagoras’ theorem to find the distance of the particle from the origin at time space t space equals space 1 spacesecond, giving your answer to three significant figures.

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6
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4 marks

The acceleration of a particle is modelled using the equation

 

bold italic a space equals space open parentheses table row cell 3 t squared space minus space 1 end cell row cell 5 e to the power of negative t end exponent end cell end table close parentheses space straight m space straight s to the power of negative 2 end exponent

 

where time t is measured in seconds.

 

(i)
Use integration to find an expression for the velocity of the particle.

(ii)
Given that when t space equals space 0 comma space bold italic v space equals spaceopen parentheses table row 4 row 0 end table close parentheses find the value of the constant(s) of integration.

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7a
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3 marks

The motion of a particle, starting from the origin, is described by the position vector

 

bold r space equals space open parentheses open parentheses 3 t cubed space minus space t close parentheses bold i space plus space open parentheses 2 t squared space minus space 1 close parentheses bold j close parentheses space straight m

 

where time t is measured in seconds.

Differentiate r with respect to t twice to find an expression for the acceleration of the particle.

7b
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3 marks
(i)
Find the acceleration of the particle at time space t space equals space 3 spaceseconds.

(ii)
Use Pythagoras’ theorem to find the magnitude of acceleration of the particle at time space t space equals space 3 spaceseconds.

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8a
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2 marks

The velocity of a particle at time t seconds is given by

 

bold v space equals space open parentheses table row cell 8 t to the power of 1 half end exponent space plus space 2 t end cell row cell 3 t squared space plus space 5 t space minus space 1 end cell end table close parentheses space space straight m space straight s to the power of negative 1 end exponent

 

Differentiate v to find the acceleration, bold a space straight m space straight s to the power of negative 2 end exponent, of the particle at time t seconds.

8b
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2 marks

Use Pythagoras’ theorem to find the magnitude of acceleration of the particle at time t space equals space 4 seconds, giving your answer to three significant figures.

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9a
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3 marks

A particle, starting from rest at the origin, has acceleration

 

bold a bold space equals space open parentheses open parentheses 6 t space minus 2 close parentheses bold i space plus space open parentheses 4 space minus space 12 t close parentheses bold j close parentheses space straight m space straight s to the power of negative 2 end exponent

 

at time t seconds.

Integrate a with respect to t twice to find an expression for the position vector of the particle. Remember to account for any constant(s) of integration.

9b
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3 marks
(i)
Find the position vector of the particle at time t equals 5 seconds.

(ii)
Use Pythagoras’ theorem to find the distance of the particle from the origin at time t equals 5 seconds, giving your answer to three significant figures.

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10a
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2 marks

The velocity of a particle at time t seconds is given by

 

bold v space equals space open parentheses open parentheses 10 t space minus space 3 t squared close parentheses bold i space plus space open parentheses 4 t space minus 5 close parentheses bold j close parentheses space straight m space straight s to the power of negative 1 end exponent

 

(i)
Differentiate v to find the acceleration, bold italic a space straight m space straight s to the power of negative 2 end exponent, of the particle at time t space seconds.

(ii)
Find the initial acceleration of the particle.
10b
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3 marks

The particle’s initial position is at the point left parenthesis space 4 space comma space 5 space right parenthesis space.

 

(i)
Integrate v to find an expression, in terms of t, for the position vector of the particle.  

            

(ii)
Find the distance of the particle from the origin at time space t space equals space 3 spaceseconds, giving your answer to three significant figures.

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11a
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1 mark

The velocity of a particle at time t seconds is given by

 

bold r with dot on top space equals space open parentheses table row cell 12 t squared space minus space 2 t end cell row cell 9 t squared space minus space 1 end cell end table close parentheses space straight m space straight s to the power of negative 1 end exponent

Differentiate bold r with dot on top to find the acceleration, bold r with bold " on top straight m space straight s to the power of negative 2 end exponent comma of the particle at time t seconds.

11b
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2 marks

Integrate bold r with dot on top to find the position vector, r m, of the particle at time t seconds given that its initial position is the origin.

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1a
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3 marks

The position of a boat on a small lake, relative to a mooring point located at the origin, is given by the vector

 bold r equals left parenthesis negative 20 space sin space left parenthesis t over 360 right parenthesis right parenthesis bold i plus left parenthesis 20 minus 20 space cos space left parenthesis t over 360 right parenthesis right parenthesis bold j space straight m

where time t is measured in seconds.

(i)
Show that the boat is initially at the mooring point.

(ii)
Show that the distance from the mooring point to the boat at time t space equals space 180 pi is 20 square root of 2 space straight m.
1b
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3 marks
(i)
Find the velocity,space space bold v space straight m space straight s to the power of negative 1 end exponent, of the boat, at time t seconds.

(ii)
Show that the boat has a speed of 1 over 18 space straight m space straight s to the power of negative 1 end exponent when it first returns to the mooring point.

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2a
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4 marks

A particle moving in the x-y  plane has velocity, bold v space straight m space straight s to the power of negative 1 end exponent, at time t seconds, given by

bold v bold space equals open parentheses table row cell 0.1 straight t cubed minus 3 straight t squared end cell row cell 2 straight t plus 1 end cell end table close parentheses

(i)
Find the acceleration, bold a bold space straight m space straight s to the power of negative 2 end exponent, of the particle at time t and explain how you can tell that the acceleration in the y-direction is constant.

(ii)
Other than t space equals space 0, find the time at which the acceleration in the x-direction is   zero.
2b
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3 marks

Find the position vector of the particle given that its initial position is at the point open parentheses negative 3 space comma space 5 close parentheses.

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3a
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3 marks

Once an aircraft reaches its cruising height (at time t = 0 hours) its acceleration is modelled by

 bold a bold space equals space open parentheses 4 t cubed space minus space 6 t squared close parentheses bold i space plus space open parentheses 0.9 t squared space minus space 1 close parentheses bold j bold space km space straight h to the power of negative 2 end exponent

Given that the velocity of the aircraft at t space equals space 5 hours is bold v space equals space 400 bold i space plus space 40 bold j bold space km space straight h to the power of negative 1 end exponent, find the velocity of the aircraft in terms of t.

3b
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2 marks

Find the speed of the aircraft when it first reaches its cruising height.

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4a
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3 marks

An ice skater moves across a straight section of a frozen river such that their position, at time t seconds relative to an origin is given by

 bold r space equals open parentheses table row cell 1 third t squared space plus space 1 fifth t end cell row cell 2 t squared space plus space 7 t end cell end table close parentheses space space straight m space space space space space space space space space space space space space space space t space greater or equal than space 0

Find the initial speed of the ice skater giving your answer to three significant figures.

4b
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3 marks

Show that the ice skater’s acceleration is constant and find the magnitude of the acceleration, giving your answer to three significant figures.

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5a
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4 marks

A remote-controlled car is driven around a large playground with velocity,  bold v subscript bold c bold space straight m space straight s to the power of negative 1 end exponent, at time t seconds, given by

bold v subscript bold c bold space equals space open parentheses 0.45 straight t squared space plus space 2 straight t space minus space 16 close parentheses bold i space plus space open parentheses 0.75 straight t squared space minus 1 close parentheses bold j

(i)
The remote-controlled car is initially set in motion from position open parentheses negative 6 space comma space 15 close parentheses. Find the position vector bold r subscript bold c of the car at time t seconds.

(ii)
Find the distance of the remote-controlled car from the origin after 15 seconds.
5b
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3 marks

At the same time as the remote-controlled car is started, a remote-controlled truck is also set into motion. The truck has position vector, rT m, at time t space seconds given by

 bold r subscript bold T equals left parenthesis 0.15 t cubed minus 6 right parenthesis bold i plus left parenthesis 0.25 t cubed minus 1 right parenthesis bold j

Determine the time(s) at which the car and the truck will collide, if at all.

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6a
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4 marks

A spider is crawling across the floor of a house such that is has acceleration

 bold a space equals space open parentheses 1.2 straight t close parentheses bold i space plus space open parentheses 0.5 close parentheses bold j space straight m space straight s to the power of negative 2 end exponent

at time t seconds after the spider emerged from under the skirting board.

 

After 3 seconds the spider’s velocity is bold v space equals space 5.4 bold i space plus 1.7 bold j space straight m space straight s to the power of negative 1 end exponent.
Find the velocity of the spider at time t seconds.

6b
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4 marks

After 3 seconds the spider’s position, relative to an origin at a corner of the floor, is left parenthesis 10.4 space comma 5.15 right parenthesis. Find the distance the spider is from the origin when it emerges from under the skirting board.

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7a
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4 marks

A stone is thrown from the edge of a deep cave such that it will fall into the cave with its motion described by the equation

  

bold v equals space open parentheses 0.2 t close parentheses bold i bold space plus open parentheses 4 space minus space 0.75 straight t squared close parentheses bold j space straight m space straight s to the power of negative 1 end exponent 

where bold v space straight m space straight s to the power of negative 1 end exponent is the velocity of the stone at time t seconds after it is thrown.

 

Find the position of the stone – relative to where it is thrown from - after 4 seconds and explain what this means in the context of the question.

7b
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3 marks

The cave has a depth of 384 m. Find the magnitude of the acceleration of the stone as it hits the bottom of the cave.

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8a
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3 marks

A particle’s velocity is modelled by the equation

bold r with. on top space equals space open parentheses table row cell 0.75 e to the power of 1.5 t end exponent space plus space 2 t end cell row cell 5 t space minus space open parentheses t space plus 1 close parentheses to the power of negative 1 end exponent end cell end table close parentheses space straight m space straight s to the power of negative 1 end exponent space space space space space space space space space space t space greater or equal than space 0

where t is the time in seconds.

The particle’s initial displacement is open parentheses table row 0 row 0 end table close parentheses.

Find the position vector of the particle, r m, at time t seconds.

8b
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3 marks

Find the magnitude of the acceleration of the particle after 1 second.

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9a
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6 marks

At time t seconds, a particle P has acceleration a m s−2, where

 bold a equals left parenthesis 4 straight t minus 3 right parenthesis bold i plus left parenthesis 4 straight t plus 5 right parenthesis bold j                              t space greater or equal than space 0                     .

Initially P starts at the origin O and moves with velocity open parentheses negative 5 bold j close parentheses space straight m space straight s to the power of negative 1 end exponent.

 

Find the distance between the origin and the position P of when t space equals space 6.

9b
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5 marks

Find the value of t at the instant when P is moving in the direction of bold i space plus space 2 bold j.

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1a
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3 marks

The position vector of a boat, sailing on a lake relative to an origin, is

 bold r equals left parenthesis 2 space sin space t space right parenthesis bold i plus left parenthesis 2 minus 2 space cos space t space right parenthesis bold j space km 

where time t is measured in hours.

(i)
Show that the boat is initially at the origin.

(ii)
Show that the boat takes hours until it first returns to the origin.
1b
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3 marks
(i)
Find the velocity, bold v space km space straight h to the power of negative 1 end exponent, of the boat, at time t hours.

(ii)
Find the velocity of the boat at time t space equals space2 over 3 straight pi hours.

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2a
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3 marks

A particle moving in a plane has velocity, bold v space straight m space straight s to the power of negative 1 end exponent, at time t seconds, given by

 

bold v space equals space open parentheses table row cell 4 t space minus space 3 t squared end cell row cell 6 t squared space minus space 2 end cell end table close parentheses 

(i)
Find an expression for the acceleration of the particle.

(ii)
Find the acceleration of the particle at time  t space equals space 3 seconds.
2b
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3 marks
(i)
Find the position vector of the particle given that its initial position is at the     

        

(ii)
Find the position vector of the particle at time space t space equals space 4 spaceseconds.

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3a
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3 marks

Once an aircraft reaches its cruising height (at time t space equals space 0 seconds) its acceleration is modelled by

 bold a space equals space open parentheses table row cell 30 minus 2 t end cell row cell 4 t space minus 3 end cell end table close parentheses space straight m space straight s to the power of negative 2 end exponent space space space space space space space space space space t greater or equal than 0

Given that the velocity of the aircraft at space t space equals space 0 spaceis bold v space equals space open parentheses table row 200 row 150 end table close parentheses space straight m space straight s to the power of negative 1 end exponent, find the velocity of the aircraft in terms of t.

3b
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3 marks

Find the speed of the aircraft at time t space equals space 4 seconds, giving your answer in kilometres per hour to three significant figures.

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4a
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4 marks

An ice skater moves across an ice rink such that their position, at time t seconds relative to an origin, is given by 

bold r space equals space open parentheses table row cell 0.2 space t to the power of 2 space end exponent minus space 0.005 t cubed end cell row cell 0.5 t space plus space 2 end cell end table close parentheses space straight m

(i)
Briefly explain how you can tell the ice skater’s motion did not start at the origin.

(ii)
Find the coordinates of the position of the ice skater after 40 seconds.

(iii)
Find the distance between the ice skater and the origin after 40 seconds.
4b
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3 marks
(i)
Find an expression for the velocity of the ice skater at time t seconds.

(ii)
Find an expression for the acceleration of the ice skater at time t seconds.

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5a
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2 marks

A remote-controlled car is driven around a playground with velocity, bold v space straight m space straight s to the power of negative 1 end exponent, at time t seconds, given by

 bold v equals left parenthesis 0.25 right parenthesis bold i plus left parenthesis 0.5 t minus 9 right parenthesis bold j 

Find an expression for the displacement of the remote-controlled car, s m, from its initial position.

5b
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1 mark

The remote-controlled car is set in motion from the point left parenthesis 2 comma space minus 5 right parenthesis. Find the position vector r of the particle at time t seconds.

5c
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4 marks
(i)
Find the distance of the remote-controlled car from its initial position after 40 seconds.

(ii)
Find the distance of the remote-controlled car from the origin after 40 space seconds.

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6a
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4 marks

A spider is crawling across the floor of a house such that is has acceleration

bold a equals left parenthesis 0.1 t right parenthesis bold i plus left parenthesis 0.6 t squared minus 2 t right parenthesis bold j space straight m space straight s to the power of negative 2 end exponent

at time t seconds after the spider emerged from under the skirting board.

 

(i)
Given that the spider’s initial velocity was  bold v equals negative 1.2 bold i plus 1.8 bold j, find the velocity of the spider at time t seconds.

(ii)
Find the speed of the spider after 3 seconds.
6b
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2 marks

The point at which the spider emerged from under the skirting board is deemed the origin. Find the position vector of the spider at time t seconds.

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7a
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2 marks

The position vector of a particle at time t seconds is given by

 bold r bold space equals space left parenthesis 12 e to the power of negative 0.1 t end exponent right parenthesis bold i bold space plus space left parenthesis 24 e to the power of negative 0.2 t end exponent right parenthesis bold j space space straight m

(i)
Write down the initial position of the particle.

(ii)
Briefly explain why the particle gets closer to the origin but never actually meets it.
7b
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2 marks

Find an expression for the velocity of the particle at time t seconds.

7c
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3 marks
(i)
Find an expression for the acceleration of the particle at time t seconds.

(ii)
Find the magnitude of acceleration of the particle at time t space equals space 4 seconds, giving your answer to three significant figures.

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8a
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2 marks

A stone is thrown from the edge of a deep cave such that it will fall into the cave with its motion described by the equation

 bold v equals space bold i plus left parenthesis 1.5 minus 0.3 t squared right parenthesis bold j space straight m space straight s to the power of negative 1 end exponent

where v is the velocity of the particle t seconds after the stone is thrown.

 

Find the speed at which the stone is thrown.

8b
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3 marks
(i)
Find the acceleration of the stone at time t seconds.

(ii)
Find the magnitude of the acceleration of the stone at time t space equals space 2.5 space seconds.

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9a
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3 marks

A particle’s velocity is modelled by the equation

 

 stack bold r bold space with dot on top space equals space open parentheses table row cell 3 t squared space minus space 6 t end cell row cell 4 space minus space 8 t cubed end cell end table close parentheses space straight m space straight s to the power of negative 1 end exponent

where t is the time in seconds.

Given that the particle is initially located at the point left parenthesis 2 space comma space 1 right parenthesis, find the position vector of the particle, r m, at time t seconds.

9b
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3 marks
(i)
Find the acceleration of the particle, bold r with bold " on top space straight m space straight s to the power of negative 2 end exponent at time t seconds.

(ii)
Find the time at which the particle has no acceleration in the horizontal direction.

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10a
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2 marks

At time t seconds, a particle P has position vector r m, where

 

                    bold r space equals space open parentheses t cubed space minus space 11 t squared space minus space 16 t space plus space 2 close parentheses bold i space plus space open parentheses t cubed space plus space 2 t space minus 1 close parentheses bold j space space space space space space space space space space space space space space space space space t space greater or equal than space 0

 

Find the velocity of P at time t seconds, wherespace t space greater or equal than space 0.

10b
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4 marks
(i)
Find the value of t at the instant when P moves in a direction parallel to j.

 

(ii)
Show that P will never move in a direction perpendicular to j.

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1a
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5 marks

The displacement of a boat, s m on a small lake, relative to a mooring point located at the point left parenthesis 10 comma space 0 right parenthesis comma is given by the vector

 bold s equals open parentheses negative 40 space sin open parentheses fraction numerator pi t over denominator 900 end fraction close parentheses close parentheses space bold i plus open parentheses 30 minus 30 space cos open parentheses fraction numerator pi t over denominator 900 end fraction close parentheses close parentheses space bold j

where time t is measured in seconds.

(i)
Write down the position vector, r m, of the boat at time t seconds.

(ii)
Find the difference between the distance the boat is from its mooring point and the distance it is from the origin at the point when t equals 225 space seconds.
1b
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5 marks
(i)
Find the velocity, bold v space straight m space straight s to the power of negative 1 end exponent, of the boat, at time t seconds.

(ii)
Given that one trip around the lake takes half an hour, find the times at which the boat is moving in one direction only during one trip around the lake.

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2
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5 marks

Once an aircraft reaches its cruising height (at time t space equals space 0 hours) its acceleration is modelled by

bold a equals left parenthesis 3 t squared minus 1 right parenthesis bold i plus left parenthesis 8 t plus 1 right parenthesis bold j space km space straight h to the power of negative 2 end exponent

Given that at time space t space equals space 2 spacehours,

  • the i-component of the aircraft’s velocity is positive and double that of the j‑component, and,
  • the speed of the aircraft is 22 square root of 5 space km space straight h to the power of negative 1 end exponent,

 find the velocity of the aircraft in terms of t.

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3a
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3 marks

An ice skater moves across a straight section of a frozen river such that their position, at time t seconds relative to an origin is given by

 bold r equals open parentheses table row cell 2 t to the power of 1 half end exponent plus t end cell row cell t to the power of 1 half end exponent end cell end table close parentheses space straight m space space space space space space space space space space space space space space space space 0 less or equal than t less or equal than 225

(i)        Find the distance the skater is from the origin after 25 seconds.


(ii)       As they skate forwards, the skater slowly crosses the width of the river.
            It takes 225 seconds for the skater to cross the river.
            How wide is the river?

3b
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4 marks

Show that the magnitude of acceleration of the skater at time t seconds is given by a equals 0.25 square root of 5 t to the power of negative 3 end exponent end root space straight m space straight s to the power of negative 2 end exponent.

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4
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8 marks

At time space t space equals space 0 spaceseconds two remote-controlled cars – one red, one blue - are started and driven around a large playground with velocities, bold space bold v subscript bold R space straight m space straight s to the power of negative 1 end exponentand bold v subscript straight B space straight m space straight s to the power of negative 1 end exponent given by

bold v subscript bold R equals left parenthesis 0.2 t minus 1 right parenthesis bold i plus left parenthesis t minus 5 right parenthesis bold j

bold v subscript bold B equals left parenthesis 0.2 t right parenthesis bold i plus left parenthesis t minus 7 right parenthesis bold j

The red car starts from position left parenthesis negative 2.4 space comma space 8 right parenthesis relative to an origin and the blue car starts from position left parenthesis negative 2.4 space comma space minus 4 right parenthesis.

 

Find the time (such that t space greater or equal than space 0) when both remoted-controlled cars are equidistant from the origin. Find the distance of the cars from the origin at this point and show that they are not in the same location.

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5
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8 marks

A spider is crawling across the floor of a house such that its acceleration is

bold a equals left parenthesis left parenthesis 0.2 t right parenthesis bold i plus left parenthesis 0.4 t right parenthesis bold j right parenthesis space straight m space straight s to the power of negative 2 end exponent

at time t seconds after the spider emerged from under the skirting board in a corner of the room.

The room measures space 12 space straight m space cross times space 8 space straight m spaceand the corner from which the spider emerges is the origin, as shown below.

edexcel-al-maths-mechanics-topic-2-4-vh---q5

The spider’s velocity is such that

  • its speed in the i-direction after 3 seconds is twice the speed in the i-direction after 2 seconds,
  • its speed in the j-direction after 3 seconds is three-times the speed in the
    j-direction after 1 second,
  • neither component of the spider’s velocity is ever negative.

 

Determine whether or not the spider is still in the room (and visible, so not under the skirting board) after 6 seconds.

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6a
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4 marks

A stone is thrown from the edge of a deep cave such that it will fall into the cave and has velocity

bold v equals left parenthesis left parenthesis 0.4 t right parenthesis bold i plus left parenthesis 2 minus 0.3 t squared right parenthesis bold j right parenthesis space straight m space straight s to the power of negative 1 end exponent

at time t seconds after it is thrown. 

Find the position of the stone at the time it is about to fall into the cave (rather than being in the air above the cave).

6b
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3 marks

Find the maximum height above the cave the stone reaches.

6c
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2 marks

The deepest known cave in the world has a depth of 2212 m. (The Veryovkina Cave in Georgia.) The model above suggests the stone would take around 28 space seconds to reach this depth. Consider the average speed and the acceleration of the stone to highlight a problem with this model.

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7a
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4 marks

A particle’s velocity is modelled by the equation

bold r with bold dot on top bold space equals space open parentheses table row cell t to the power of 1 half end exponent minus t end cell row cell 4 left parenthesis t plus 1 right parenthesis to the power of negative 1 end exponent plus 5 t to the power of 3 over 2 end exponent end cell end table close parentheses space straight m space straight s to the power of negative 1 space end exponent space space space space space space space space space space space t greater or equal than 0

where t is the time in seconds.

 

The particle’s initial position is space left parenthesis 3 space comma space 5 right parenthesis comma spacefind the position vector of the particle, r m, at time t seconds.

7b
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3 marks

Find the time at which the particle’s acceleration, bold r with bold " on top bold space straight m space straight s to the power of negative 2 end exponent is zero in the horizontal (i) direction.

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8a
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3 marks

The position vectors in the x-y plane of two particles, A and B, at time t, left parenthesis t space greater or equal than space 0 right parenthesis are given by

 bold r subscript bold A equals left parenthesis 3 e to the power of negative 0.15 t end exponent right parenthesis bold i plus left parenthesis 4 e to the power of 0.1 t end exponent right parenthesis bold j

bold r subscript bold B equals left parenthesis negative 3 e to the power of negative 0.15 t end exponent right parenthesis bold i plus left parenthesis 4 e to the power of 0.1 t end exponent right parenthesis bold j

(i)
Write down the initial position of both particles.

(ii)
Briefly explain what happens to the position of both particles for very high values of t.
8b
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3 marks

(i)        Find the velocity of particle A in terms of t.

(ii)       Hence write down the velocity of particle B in terms of t.

8c
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3 marks

(i)        On the same diagram sketch the graphs of y against x for both rA and  rB.


(ii)       Without doing any calculations explain why for all values of t,

open vertical bar bold r subscript bold A close vertical bar space equals space open vertical bar bold r subscript bold B close vertical bar comma space open vertical bar bold v subscript bold A close vertical bar space equals space open vertical bar bold v subscript bold B close vertical bar space and space open vertical bar bold a subscript bold A close vertical bar space equals space open vertical bar bold a subscript bold B close vertical bar

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