Chapter 9 - Motion Along a Straight Lines

 Additional Mathematics Module Form 5 SMK Agama Arau, Perlis

Chapter 9- Motion Along A Straight Line

CHAPTER 9 - MOTION ALONG A STRAIGHT LINE 9.1 DISPLACEMENT

- Displacement, s m, from a fixed point O is a function of time  f(t). Example:

s

=

t 3

−

11 2

t 2

+

6t  or  f (t ) = t 3

−

11 2

t 2

6t 

+

- Positive displacement, s > 0 , is when the particle is on the positive side of O. - Negative displacement, s < 0 , is when the particle is on the negative side of O. - Zero displacement, s = 0 , is when the particle is at the point O. - When the displacement is maximum or minimum,

ds dt 

=

0.

2

- The displacement is maximum when

d  s 2

dt 

2

<

0 and minimum when

d  s 2

dt 

>

0

Example 1:

A particle Q moves along a straight line from a fixed point O such that i ts displacement, s m, t seconds after passing through O is given by s

=

t 3

−

11 2

t 2

+

6t .

Find: (a)The displacement of Q at t =2 s (b) The distance travelled in the third second. (c) the time t , when Q is instantaneously at rest. (d) The displacement of Q when Q is instantaneously at rest for the second time. v  = 6ms

Solution:

=

( 2)

= −

1

s=0

(a) t = 2 s,

s

−

3

11

−

2

2m ( 2)

2

+

6( 2)

t  = 3

2m

t − 0

O

t  = 2 4.5m

(b) t = 3 s,

s

=

(3) 3

−

11 2

(3) 2

+

6(3)

4.5m

= −

Distance travelled in the third second is 4.5 – 2

Distance travelled in third second means the distance travelled from 2

nd

rd

second to 3 second

= 2.5 m

Page | 1

 Additional Mathematics Module Form 5 SMK Agama Arau, Perlis

Chapter 9- Motion Along A Straight Line

(c) Instantaneously at rest, v= 0

s

=

ds dt  ds dt 

3

t 

−

11 2

=

3t 2

=

v  2

v  = 3t 

+

6t  Differentiate with respect to t 

−

−

2

t 

11t + 6

11t  + 6

v  = 0, 3t 2

−

11t + 6 = 0

(3t  − 2)(t  − 3) = 0 t  =

2

or t  = 3

3

Time t , when Q is instantaneously at rest i s t  =

2 3

s or t  = 3s .

(d ) The displacement of Q when Q is instantaneously at rest for the second time is when t  = 3s ,

s

=

(3)

3

−

11 2

(3)

4.5m

= −

2

+

6(3) ‘negative’ means opposite direction, not magnitude

9.2 VELOCITY -1

- Velocity, v ms , at a given instant is the rate of change of displacement or v  =

ds dt 

.

Example: 2

v  = 3t 

−

11t + 6

- Positive displacement, v > 0 , is when the particle is moving to the right. - Negative displacement, v < 0 , is when the particle is moving to the left. - Zero displacement, v = 0 , is when the particle is not moving or temporarily at rest. - Displacement is maximum when the velocity is zero.

- Velocity is maximum or minimum when acceleration is zero. - Given the velocity function v = f(t), therefore the displacement function, s

=

∫  f (t )dt 

Page | 2

 Additional Mathematics Module Form 5 SMK Agama Arau, Perlis

Chapter 9- Motion Along A Straight Line

Example 2: -1

A particle Q moves along a straight such that its velocity, v ms , t seconds after passing through O is 3

given by v  = 2t 

−

2

5t 

+

4t + 6 .

(a) Find the initial velocity of Q  (b) Find the acceleration when t = 2 (c) If the acceleration of Q is zero when t = a and t = b , find the values of  a and b (d) Find the displacement of P in terms of  t  Solution:

(a) Initial velocity = 6 ms 3

(b) v  = 2t 

−

dv 

3

6t 

=

dt 

2

a = 6t 

5t 2 −

−

+

-1

Initial velocity is when t = 0

4t + 6 Differentiate with respect to t.

10t  + 4

a

10t + 4

=

dv  dt 

t  = 2, 2

a = 6( 2) 8m

= 2

(c) 6t 

− 3

3t 

−

10( 2) + 4

2

−

10t + 4 = 0

−

5t  + 2 = 0

(3t  − 2)(t − 1) 2

t  = a

2

or b = 1

3 3

(d) v  = 2t 

−

∫ 

3

s = ( 2t  4

=

t 

2

0

or t  = 1

3

=

=

5t 2

+ 2

5t 

−

+

3

−

4

s

=

2

∫ 

s = v ..dt 

4t  + 6)dt 

5t  3

+

2

2t 

+

6t + c

Substitute s = 0, t = 0 into

t 

Integrate with respect respect to t.

4t + 6

1

1

,

3

−

5t  3

+

2

2t 

+

6t 

Page | 3

 Additional Mathematics Module Form 5 SMK Agama Arau, Perlis

Chapter 9- Motion Along A Straight Line

9.3 ACCELERATION -2

- Acceleration, a ms , is defined as the rate of change of velocity with time or a

=

dv  dt 

2

or a

=

d  s 2

dt 

Example:

a

6t  − 11

=

- Positive acceleration, a > 0 , is when the particle is accelerating. - Negative acceleration, a < 0 , is when the particle is decelerating - Zero acceleration, a = 0 , is when the particle is moving with uniform velocity.

∫ 

- Given the acceleration function function a = f(t), therefore the velocity function, v  = a..dt 

Example 3: -1

A particle Q moves along a straight line from a fixed point O such that i ts velocity, v ms , t seconds after 3

passing through O is given by v  = t 

−

3t 2 .

Find: (a) The acceleration of Q when Q is instantaneously at rest after passing through O. (b) the interval of  t , when Q is decelerating.

Solution: (a) v= 0 ,

t 3

−

(b) Q is decelerating, a < 0,

2 3t 

=

2

t  (t  − 3)

0

3t 

0

t 

=

t  = 0 or  t  = 3 v  = t 3 dv  dt  a

=

=

−

2

3t 

−

−

2

−

−

6t  < 0

3t  < 0

t (t  − 2)

<

0

let  t (t − 2) = 0

3t 2

2 3t 

2

6t 

0 2) < t  t t (=t − 0= 2

6t 

t  = 3 ,

a = 3(3)

2

9ms

−

=

−

6(3)

2

The interval of t is

0 < t  < 2 Page | 4

 Additional Mathematics Module Form 5 SMK Agama Arau, Perlis

Chapter 9- Motion Along A Straight Line

CHAPTER REVIEW EXERCISE  –1

1. A particle moves along a straight line and passes through a fixed point O. Its velocity, v m s , is given 2

by v = pt  + qt – 16 , where t is the time, in seconds, after passing through O, p and q are constants. The particle stops momentarily at a point 64 m to the left of  O when t = 4. [Assume motion to the right is positive] (a) Find the initial velocity of the particle,  p and of q, (b) Find the value of  p

(c) Find the acceleration of the particle when it stops momentarily, (d) Find the total distance traveled in the third second. -1 2 2. A particle moves along a straight straight line from a fixed point point O. Its velocity, v ms , is given by v = 15t  –  – 3t  ,

where t is the time, in i n seconds, after leaving the point O. (Assume motion to the right is positive) (a) Find the maximum velocity of the particle, (b) Find the distance travelled during the fourth second, (c) Find the value of  t when the particle passes the point O again, (d) Find the time between leaving O and when the particle reverses its direction of motion. -1

3. A particle moves along a straight line and passes a fixed point O, with a velocity of 10 ms . Its -2

acceleration, a ms , t s after passing through O is given by a = 2t  –  – 7. (Take the direction to the right as the positive direction) (a) Find the constant velocity of the particle. (b) Find the range of time tim e for which the particle moves to the left. (c) Find the total distance travelled by the particle in the first 5 seconds. -1 4. A particle moves along a straight line and passed through a fixed point O. Its velocity, v ms , is given 2

by v = t  − 5t + 4 , where t is the time, in i n seconds, after passing through O. [Assume motion to the right is positive.] -1

(a) Find the initial velocity, in m s , -1

(b) Find the maximum velocity, in m s , (c) Find the range of time when particle moves to the left, (d) Find the total distance, in m , traveled by the particle in the first four seconds. -1 5. A particle moves along a straight line and passes through a fixed point P. Its v elocity, v ms , is given by

v = 4t – 8, where t is the time, in seconds, after af ter passing through the point P.

(Assume motion to the right is positive) -1

(a) Find the initial velocity, in ms ,of the particle. (b) Find the value of  t when the particle stops instantaneously,  – 1

(c) Find the displacement of the particle from P when its velocity is – 4 ms

.

(d) Find the total distance travelled, in m, by the particle during the first 4 seconds.

Page | 5