M2 Particle kinematics - KUMAR'S MATHS REVISION · The constant acceleration equations were used...

M2 PARTICLE KINEMATICS http://kumarmaths.weebly.com

1

Differentiating and Integrating Vectors

The constant acceleration equations were used extensively in M1 but in M2

displacement is a function of time. It follows then that velocity, which is

the rate of change of displacement, can be found by differentiating the

displacement function.

Displacement

Differentiate Velocity Integrate

(Remember the constant)

Acceleration

If r is the position vector, and using dot notation: -

.

2..

2

drr velocity

dt

d rr acceleration

dt

Example 1

The position vector of a particle Q at time t is 2(3 1) 2r t i t j

(r measured in metres).

Find the initial position vector and show that the acceleration is constant

Initial position is when t=0

r i

Remembering that ..

r acceleration

.

3 4r i tj

..

4r j Since the acceleration vector has no variable t it is said to be constant.

2

Integrating Vectors

Since a vector has both i and j components it will also have two separate

functions with respect to time.

Therefore if ..

( ) ( )r f t i g t j

We need to integrate each function separately, remembering the constants.

Therefore

.

1 2( ) ( )r i f t dt j g t dt C i C j

Where 1 2,C C are constants of integration.

R is found by integrating the velocity function with respect to time.

Example 2

A particle moves such that at time t

.

24 5r ti t At time t = 0 the particle has a position vector 5i – 6j Find the position vector at time t.

The position vector is found by integrating the velocity vector.

21 2

2 31 2

4 5

52

3

r i tdt j t dt C i C j

r t i t j C i C j

At time t = 0, r = 5i – 6j 1 5C , 2 6C

2 35(2 5) 6

3r t i t j

3

Example 3

A particle Q has position vector ( 25i – 40j )m at time t = 0 relative to the

origin. Q moves with constant acceleration and is equal to -2 ( 5i + 12j )ms .

When t = 0, .

0r . Find:

a) .

r when t = 4

b) The distance of Q from O at this time.

a) By integrating the acceleration vector we can find the velocity vector.

..

.

1 2

.

1 2

5 12

5 12

5 12

r i j

r i dt j dt C i C j

r ti tj C i C j

At time t = 0, .

0r 1 2 0C i C j

.

5 12r ti tj

When t = 4 .

1(20 48 ) r i j ms

b) To find the distance OQ we need the position vector (r)

.

5 12r ti tj

1 2

2 21 2

5 12

56

2

r i tdt j tdt K i K j

r t i t j K i K j

When t = 0, r = 25i – 40j

2 25( 25) (6 40)2

r t i t j

Substituting t = 4 gives 65 56 r i j

Distance OQ is the magnitude of r

85.8

OQ r

m

4

Example 4

A remote control car is being tested in a horizontal playground. At time t

seconds, the position vector, r metres, of the car relative to a fixed point O

is given by

52 2

9 8

2 5r t i t j

At the instant when t = 4,

a) show that the car is moving with velocity -1(36i + 32j)ms

b) find the magnitude of the acceleration of the car.

A cyclist is moving with constant velocity -117jms . At the instant when t = 4,

calculate

c) the velocity of the car relative to the cyclist;

d) the speed of the car relative to the cyclist;

e) the acute angle between the relative velocity and the constant velocity of

the cyclist.

a)

drVelocity

dt

52 2

3

2

9 8

2 5

9 4

r t i t j

v ti t j

When t = 4 1(36 32 )Velocity i j ms

b)

dvAcceleration

dt

1

29 6dv

i t jdt

When t = 4 9 12a i j

So

2 2

2

(9 12 )

15

a

a ms

c) The velocity of the car relative to the cyclist is given by:

36 32 17 36 15car cyclistv v i j j i j

5

d) the speed of the car relative to the cyclist is given by:

2 2(36 15 )car cyclistv v

139ms

e) The cyclist is moving parallel to the vector j.

The car has velocity1(36 32 ) i j ms which can be represented

diagrammatically as:

The required angle is θ, where 36

tan15

67.4

Example 5

A particle P of mass 5kg is acted on by a constant force F. At time t

seconds the position of the particle, r metres, is given by the equation

2 2(3 2) ( 4 )r t kt i kt t k j

where k is a positive constant.

a) Find the acceleration of P in 2ms in terms of k

b) Given that the magnitude of F is 50N, calculate the value of k.

The particle is moving in a direction parallel to j when t = T.

c) Find the value of T.

d) Hence find, to the nearest 0.1º, the angle that the position vector

makes with the direction of i when t = T

j

θ

θ 36i+15j

x

y

6

a) To find the acceleration we need to differentiate the position vector

twice.

2 2

2

2

(3 2) ( 4 )

(6 ) (2 4)

6 2

r t kt i kt t k j

drt k i kt j

dtd r

i kjdt

b) F = 50N and by using F = ma

2 2 2

2

2

2

5(6 2 )

(30 10 )

10 (9 )

50

5 (9 )

16

4

F i kj

F k

F k

F

k

K

K

But K is positive, so K = 4

c) The particle is now traveling parallel to j. Therefore the i component of

the velocity must equate to zero.

(6 ) (2 4) v t k i kt j Therefore:

6 4 0

2

3

t

t

d) When

2

3t

2 2(3 4 2) (4 4 ) r t t i t t k j

2 4

3 9 r i j

The angle that the position vector makes with the vector i is given by: 4

9

2

3

tan

33.7

M2 Particle Kinematics Page 7

Questions B

1 At time t=0 a particle Q is at the point with position vector (6i + 10j)m

relative to a fixed origin O. The particle moves with constant acceleration ..

r

where ..

2(5 12 ) r i j ms . Given that when t=0, .

0r find

a) the velocity, .

r , when t=3,

b) the distance of Q from O at this time.

2 A particle Q moves such that at time t seconds, t≥0, its position vector, r

metres, relative to a fixed origin is given by

3 2(3 2) ( 2 ) r t t i t t j

a) Find the velocity of Q when t = 3.

The velocity of Q is parallel to (3i – j) when t = T.

b) Find the value of T.

3 A particle moves so that at time t seconds its position vector, rm, relative

to a fixed origin is given by:

2 3 2( 4 ) ( ) r t t i t bt j where b is a constant.

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

b) Given that the particle comes to instantaneous rest, find the value of b.

4 A ball of mass 0.1kg is hit by a bat which gives it an impulse of

( 3.5i + 3j)Ns. The velocity of the ball immediately after being hit is 1(10 25 ) i j ms .

a) Find the velocity of the ball immediately before it was hit.

In the subsequent motion the ball is modeled as a particle moving freely under

gravity. When it is hit the ball is 0.8m above the ground.

b) Find the greatest height of the ball above the ground.

The ball is caught when it is 0.8m above the ground again.

c) Find the distance from the point where the ball is hit to the point

where it is caught. June 2001, Q1.

At time t seconds, a particle P has position vector r metres relative to a fixed origin O, where

r = (t2 + 2t)i + (t 2t2)j.

Show that the acceleration of P is constant and find its magnitude.

(5)


1. A particle P of mass 0.3 kg is moving under the action of a single force F newtons. At time t

seconds the velocity of P, v m s1, is given by

v = 3t2i + (6t – 4)j.

(a) Calculate, to 3 significant figures, the magnitude of F when t = 2.

(5)

When t = 0, P is at the point A. The position vector of A with respect to a fixed origin O is (3i

– 4j) m. When t = 4, P is at the point B.

(b) Find the position vector of B.

(5)

Q3, Jan 2002

2. The velocity v m s1 of a particle P at time t seconds is given by

v = (3t 2)i 5tj.

(a) Show that the acceleration of P is constant.

(2)

At t = 0, the position vector of P relative to a fixed origin O is 3i m.

(b) Find the distance of P from O when t = 2.

(6)

Q1, June 2002

3. A particle P moves in a straight line so that, at time t seconds, its acceleration a m s2 is given

by

.3,27

,30,4

2

2

tt

ttt

a

At t = 0, P is at rest. Find the speed of P when

(a) t = 3,

(3)

(b) t = 6.

(5)

Q2, June 2002

4. A particle P moves on the x-axis. The acceleration of P at time t seconds is (4t – 8) m s2,

measured in the direction of x increasing. The velocity of P at time t seconds is v m s1. Given

that v = 6 when t = 0, find

(a) v in terms of t,

(4)

(b) the distance between the two points where P is instantaneously at rest.

(7)

Q5, Jan 2003


5. A particle P moves on the x-axis. At time t seconds the velocity of P is v m s1 in the direction

of x increasing, where v = 6t – 2t2. When 0t , P is at the origin O. Find the distance of P from

O when P comes to instantaneous rest after leaving O.

(5)


(4)

(b) the distance between the two points where P is instantaneously at rest.

(7)

Q1, June 2003

6. A particle P of mass 0.75 kg is moving under the action of a single force F newtons. At time

t seconds, the velocity v m s1 of P is given by

v = (t2 + 2)i 6tj.

(a) Find the magnitude of F when t = 4.

(5)

When t = 5, the particle P receives an impulse of magnitude 92 Ns in the direction of the

vector i j.

(b) Find the velocity of P immediately after the impulse.

(4)

Q2, Jan 2004

7. At time t seconds, the velocity of a particle P is [(4t 7)i 5j] m s1. When t = 0, P is at the

point with position vector (3i + 5j) m relative to a fixed origin O.

(a) Find an expression for the position vector of P after t seconds, giving your answer in the

form (ai + bj) m.

(4)

A second particle Q moves with constant velocity (2i 3j) m s1. When t = 0, the position

vector of Q is (7i) m.

(b) Prove that P and Q collide.

(6)

Q4, June 2004



t seconds, the velocity of P, v m s–1, is given by

v = (6t + 4)i + (t2 + 3t)j.

When t = 0, P is at the point with position vector (–3i + 4j) m. When t = 4, P is at the point S.

(a) Calculate the magnitude of F when t = 4.

(4)

(b) Calculate the distance OS.

(5)

Q4, Jan 2005

9. A particle P moves in a horizontal plane. At time t seconds, the position vector of P is r metres

relative to a fixed origin O, and r is given by

r = (18t – 4t

3)i + ct

2j,

where c is a positive constant. When t = 1.5, the speed of P is 15 m s–1. Find

(a) the value of c,

(6)

(b) the acceleration of P when t = 1.5.

(3)

Q3, June 2005

10. A particle P of mass 0.4 kg is moving so that its position vector r metres at time t seconds is

given by

r = (t2 + 4t)i + (3t – t3)j.

(a) Calculate the speed of P when t = 3.

(5)

When t = 3, the particle P is given an impulse (8i – 12j) N s.

(b) Find the velocity of P immediately after the impulse.

(3)

Q2, Jan 2006

11. A particle P moves on the x-axis. At time t seconds, its acceleration is (5 – 2t) m s–2, measured

in the direction of x increasing. When t = 0, its velocity is 6 m s–1 measured in the direction

of x increasing. Find the time when P is instantaneously at rest in the subsequent motion.

(6)

Q1, June 2006


12. A cricket ball of mass 0.5 kg is struck by a bat. Immediately before being struck, the velocity of

the ball is (–30i) m s–1. Immediately after being struck, the velocity of the ball is (16i + 20j) m

s–1 .

(a) Find the magnitude of the impulse exerted on the ball by the bat.

(4)

In the subsequent motion, the position vector of the ball is r metres at time t seconds. In a

model of the situation, it is assumed that r = [16ti + (20t – 5t2)j]. Using this model,

(b) find the speed of the ball when t = 3.

(4)

Q3, June 2006


t seconds, F = (1.5t 2 – 3)i + 2tj. When t = 2, the velocity of P is 1( 4 5 )ms . i j

(a) Find the acceleration of P at time t seconds.

(2)

(b) Show that, when t = 3, the velocity of P is 1(9 15 )ms .i j

(5)

When t = 3, the particle P receives an impulse Q N s. Immediately after the impulse the velocity

of P is 1( 3 20 )ms . i j Find

(c) the magnitude of Q,

(3)

(d) the angle between Q and i.

(3)

Q6, Jan 2007

14. A particle P of mass 0.5 kg moves under the action of a single force F newtons. At time

t seconds, the velocity v m s–1 of P is given by

v = 3t2i + (1 – 4t)j.

Find

(a) the acceleration of P at time t seconds,

(2)

(b) the magnitude of F when t = 2.

(4)

Q2, June 2007


15. A particle P moves on the x-axis. At time t seconds the velocity of P is v m s–1 in the direction

of x increasing, where v is given by

.4,216

40,8 2

23

tt

tttv

When t = 0, P is at the origin O.

Find

(a) the greatest speed of P in the interval 0 t 4,

(4)

(b) the distance of P from O when t = 4,

(3)

(c) the time at which P is instantaneously at rest for t > 4,

(1)

(d) the total distance travelled by P in the first 10 s of its motion.

(8)

Q8, June 2007

16. At time t seconds (t 0), a particle P has position vector p metres, with respect to a fixed origin

O, where

p = (3t2 – 6t + 4)i + (3t3 – 4t)j.

Find

(a) the velocity of P at time t seconds,

(2)

(b) the value of t when P is moving parallel to the vector i.

(3)

When t = 1, the particle P receives an impulse of (2i – 6j) N s. Given that the mass of P

is 0.5 kg,

(c) find the velocity of P immediately after the impulse.

(4)

Q2, Jan 2008



t seconds,

F = (6t – 5) i + (t2 – 2t) j.

The velocity of P at time t seconds is v m s–1. When t = 0, v = i – 4j.

(a) Find v at time t seconds.

(6)

When t = 3, the particle P receives an impulse (–5i + 12j) N s.

(b) Find the speed of P immediately after it receives the impulse.

(6)

Q4, May 2008

18. A particle P moves along the x-axis in a straight line so that, at time t seconds, the velocity of P

is v m s–1, where

v =

.6,432

,60,210

2

2

tt

ttt

At t = 0, P is at the origin O. Find the displacement of P from O when

(a) t = 6,

(3)

(b) t = 10.

(5)

Q4, Jan 2009

19. A particle of mass 0.25 kg is moving with velocity (3i + 7j) m s–1 when it receives the impulse

(5i – 3j) N s.

Find the speed of the particle immediately after the impulse.

(5)

Q1, May 2009

20. At time t = 0 a particle P leaves the origin O and moves along the x-axis. At time t seconds the

velocity of P is v m s–1, where

v = 8t − t 2 .

(a) Find the maximum value of v.

(4)

(b) Find the time taken for P to return to O.

(5)

Q2, May 2009


21. A particle P moves along the x-axis. At time t seconds the velocity of P is v m s–1 in the positive

x-direction, where v = 3t2 – 4t + 3. When t = 0, P is at the origin O. Find the distance of P from

O when P is moving with minimum velocity.

(8)

Q1, Jan 2010

22. A particle P moves on the x-axis. The acceleration of P at time t seconds, t ≥ 0, is (3t + 5) m s–

2 in the positive x-direction. When t = 0, the velocity of P is 2 m s–1 in the positive x-direction.

When t = T, the velocity of P is 6 m s–1 in the positive x-direction.

Find the value of T.

(6)

Q1, June 2010 23. A particle moves along the x-axis. At time t = 0 the particle passes through the origin with

speed 8 m s–1 in the positive x-direction. The acceleration of the particle at time t seconds, t 0,

is (4t3 – 12t) m s–2 in the positive x-direction.

Find

(a) the velocity of the particle at time t seconds,

(3)

(b) the displacement of the particle from the origin at time t seconds,

(2)

(c) the values of t at which the particle is instantaneously at rest.

(3)

Q3, Jan 2011

24. A particle P moves on the x-axis. The acceleration of P at time t seconds is (t − 4) m s−2 in the

positive x-direction. The velocity of P at time t seconds is v m s–1. When t = 0, v = 6.

Find


(4)

(b) the values of t when P is instantaneously at rest,

(3)

(c) the distance between the two points at which P is instantaneously at rest.

(4)

Q6, June 2011

25. A tennis ball of mass 0.1 kg is hit by a racquet. Immediately before being hit, the ball has

velocity 30i m s−1. The racquet exerts an impulse of (−2i − 4j) N s on the ball. By modelling

the ball as a particle, find the velocity of the ball immediately after being hit.

(4)

Q1, Jan 2012


26. A particle P is moving in a plane. At time t seconds, P is moving with velocity v m s−1, where v

= 2ti − 3t2 j.

Find

(a) the speed of P when t = 4,

(2)

(b) the acceleration of P when t = 4.

(3)

Given that P is at the point with position vector (−4i + j) m when t = 1,

(c) find the position vector of P when t = 4.

(5)

Q2, Jan 2012

27. [In this question i and j are perpendicular unit vectors in a horizontal plane.]

A particle P moves in such a way that its velocity v m s–1 at time t seconds is given by

v = (3t 2 – 1)i + (4t – t 2)j.

(a) Find the magnitude of the acceleration of P when t = 1.

(5)

Given that, when t = 0, the position vector of P is i metres,

(b) find the position vector of P when t = 3.

(5)

Q1, May 2012


of x increasing, where

v = 2t2 – 14t + 20, t ≥ 0

Find

(a) the times when P is instantaneously at rest,

(3)

(b) the greatest speed of P in the interval 0 ≤ t ≤ 4,

(5)

(c) the total distance travelled by P in the interval 0 ≤ t ≤ 4.

(5)

Q3, June 2013


29. A particle P moves along a straight line in such a way that at time t seconds its velocity

v m s–1 is given by

213 4

2v t t

Find

(a) the times when P is at rest,

(4)

(b) the total distance travelled by P between t = 0 and t = 4. (5)

Q3, June 2013_R

30. At time t seconds, where t ≥ 0, a particle P is moving on a horizontal plane with acceleration

[(3t2 – 4t)i + (6t – 5)j] m s–2.

When t = 3 the velocity of P is (11i + 10j) m s–1.

Find

(a) the velocity of P at time t seconds,

(5)

(b) the speed of P when it is moving parallel to the vector i.

(4)

Q2, June 2014

31. A particle P moves on the positive x-axis. The velocity of P at time t seconds is (2t2

– 9t + 4) m s–1. When t = 0, P is 15 m from the origin O.

Find

(a) the values of t when P is instantaneously at rest,

(3)

(b) the acceleration of P when t = 5,

(3)

(c) the total distance travelled by P in the interval 0 t 5.

(5)

Q6, June 2015


32. A particle P moves along a straight line. The speed of P at time t seconds (t ≥ 0) is v m s–1,

where v = (pt2 + qt + r) and p, q and r are constants. When t = 2 the speed of P has its minimum

value. When t = 0, v = 11 and when t = 2 , v = 3.

Find

(a) the acceleration of P when t = 3,

(8)

(b) the distance travelled by P in the third second of the motion.

(5)

Q1, June 2016


of x increasing, where

v = (t – 2)(3t – 10), t ≥ 0

When t = 0, P is at the origin O.

(a) Find the acceleration of P when t = 3.

(3)

(b) Find the total distance travelled by P in the first 3 seconds of its motion.

(6)

(c) Show that P never returns to O.

(2)

Q2, IAL Jan 2014

34. A particle P moves on the x-axis. The acceleration of P, in the positive x direction at time

t seconds, is (2t – 3) m s–2. The velocity of P, in the positive x direction at time t seconds, is

v m s–1. When t = 0, v = 2.

(a) Find v in terms of t.

(4)

The particle is instantaneously at rest at times t1 seconds and t2 seconds, where t1 < t2.

(b) Find the values t1 and t2.

(3)

(c) Find the distance travelled by P between t = t1 and t = t2.

(4)

Q1, IAL June 2014


35. At time t seconds (t ≥ 0) a particle P has position vector r metres, with respect to a fixed origin

O, where

4 2 212 5 (5 )

8t t t t

r i j

and λ is a constant.

When t = 4, P is moving parallel to the vector j.

(a) Show that λ = 2.

(5)

(b) Find the speed of P when t = 4.

(1)

(c) Find the acceleration of P when t = 4.

(2)

When t = 0, P is at the point A. When t = 4, P is at the point B.

(d) Find the distance AB.

(4)

Q3, IAL Jan 2015

36. At time t seconds, t ≥ 0, a particle P has velocity v m s–1, where

v = (27 – 3t2)i + (8 – t3)j

When t = 1, the particle P is at the point with position vector r m relative to a fixed origin O,

where r = –5i + 2j.

Find

(a) the magnitude of the acceleration of P at the instant when it is moving in the direction of

the vector i,

(5)

(b) the position vector of P at the instant when t = 3.

(5)

Q2, IAL June 2015


37. At time t seconds (t ≥ 0) a particle P has velocity v m s–1, where

v = (6t2 + 6t)i + (3t2 + 24)j

When t = 0 the particle P is at the origin O. At time T seconds, P is at the point A and

v = λ(i + j), where λ is a constant.

Find

(a) the value of T,

(3)

(b) the acceleration of P as it passes through the point A,

(3)

(c) the distance OA.

(5)

Q3, IAL Jan 2016

38. A particle of mass 3 kg is moving with velocity (3i + 5j) m s–1 when it receives an impulse

(–4i + 3j) N s.

Find

(a) the speed of the particle immediately after receiving the impulse,

(5)

(b) the kinetic energy gained by the particle as a result of the impulse.

(3)

Q1, IAL June 2016

39. At time t seconds (t ≥ 0), a particle P has position vector r metres with respect to a fixed

origin O, where

r = 3 2924

2t t t

i + 3 23 12t t t j

At time T seconds, P is moving in a direction parallel to the vector –i – j

Find

(a) the value of T,

(5)

(b) the magnitude of the acceleration of P at the instant when t = T.

(5)

Q4, IAL Oct 2016


40. A particle P moves along a straight line. At time t = 0, P passes the point A on the line

and at time t seconds the velocity of P is v m s–1 where

v = (2t – 3)(t – 2)

At t = 3, P reaches the point B. Find the total distance moved by P as it travels from

A to B.

(6)

Q3, IAL Jan 2017

M2 Particle kinematics - KUMAR'S MATHS REVISION · The constant acceleration equations were used...

Documents

Transcript of M2 Particle kinematics - KUMAR'S MATHS REVISION · The constant acceleration equations were used...