Motion Along A Straight Line

A - DISPLACEMENT

1 The distance of a particle measured in a particular direction from a fixed
point is called the displacement of the particle from that point.

2 When a particle moves along a straight line, its displacement from a fixed
point changes with time. Hence, the displacement of a particle is a
function of time, ie s = f(t), where s is the displacement and t is the time.

3 Displacement is a vector quantity. If O is the point of reference,
(a) the displacement is positive if the particle is at a distance to the
right of the reference point.
(b) the displacement is negative if the particle is at a distance to the
left of the reference point.
(c) the displacement is zero if the particle is at point O (passing through
O or returns to O).

s negative s positive

O
s=0

4 The total distance travelled by a particle in the first n seconds is the
distance travelled from the time when t = 0 to the time when t = n sec.
For example, the first 5 seconds means from the time t = 0 to t = 5 sec.

5 The distance travelled in nth second is the distance travelled from the
time when t =n - 1 second to the time when t = n second.
For example, the 7th second means from the time t = 6 sec to t = 7 sec.

Distance travelled in the nth second = l sn - sn-1 l.

Example 1

A particle moves along a straight line. The displacement, s m, from a fixed
point O at t sec is given by s = 25t - 5t 2.
(a) Find the displacement of the particle when

(i) t = 1 sec,
(ii) t = 8 sec.
(b) Find the time when the particle
(i) is 30 m to the right of point O,
(ii) is 30m to the left of point O,
(iii) passes through the point O again.

(a) (i) t = 1 sec, s = 25(1) - 5(12) = 20 m.

(ii) t = 8 sec, s = 25(8) - 5(82) = -120 m.

(b) (i) s = 30m, 30 = 25t - 5t2
5t2- 25t + 30 = 0
t2- 5t + 6 = 0

(t - 3) (t - 2) = 0
t = 3 sec or t = 2 sec

(ii) s = -30m, -30 = 25t - 5t2
5t2- 25t - 30 = 0
t2- 5t - 6 = 0

(t - 6) (t + 1) = 0
t = -1 sec (rejected)
t = 6 sec

(iii) s = 0m, 0 = 25t - 5t2
0 = 5t - t2
0 = t(5 - t)
t = 0 (initial position of particle at point O)
t = 5 sec (the time when particle returns back to O again)

Example 2

The displacement, s m, of a particle that moves along a straight line is given by
s = t2 - 8t + 15, where t is the time in seconds after passing through point O.
Find the range of time, t, when the particle is at a distance to the
(a) right of point O,
(b) left of point O.

(a) t2 - 8t + 15 > 0 35 t
(t - 3) (t - 5) > 0 t > 5 or 0 ≤ t < 3

(b) t2 - 8t + 15 < 0
3<t<5

Example 3

A particle is moving along a straight line. Its displacement, s m, from a fixed
point O at time t seconds is given by s = 4t - t2 + 12.
Find (a) the total distance travelled by the particle in the first 6 seconds,

(b) the distance covered by the particle in the 7th second.

s
(a) 16

12

26 t

s = 0, 0 = t2 - 4t - 12
0 = (t - 6) (t + 2)
t=6


= 4 − 2 = 0

2 = 4

= 2
ℎ = 2, = 4 2 − 22 + 12 = 16

Hence, total distance travelled in the first 6 seconds (from t = 0 to t = 6)
= (16 - 12) + 16
= 20 m

(b) t = 6, s = 4(6) -62 + 12 = 0

t = 7, s = 4(7) -72 + 12 = -9

Distance covered in the 7th second (from t = 6 to t = 7)
= |s7 - s6|
= |-9 - 0|
=9m

B - VELOCITY

1 The velocity of a particle moving along a straight line is the rate of change
of displacement with time.

2 Displacement, s, is a function of time, t, or s = f(t). By differentiating s with
respect to t,

, = .

3 The velocity, v, of a particle moving along a straight line at a particular
instant is its instantaneous velocity.

4 If the motion of a particle towards the right is taken as positive direction,
then
(a) a negative velocity means that the particle is moving from
right to left,
(b) a positive velocity means that the particle is moving from
left to right,
(c) a zero velocity means that the particle is stationary.

v - v+ v=0 v- v+ v-
s+ s+
s - s- s-
O
s=0

5 When a particle is at instantaneous rest, its direction is reversed.

6 Since v(t) = ds/dt, , = න .

7

= න .

−

8 The displacement is maximum (at the right-hand side or the left-hand side
of a fixed point) when the velocity is zero, ie ds/dt = 0.

9 The total distance travelled by a particle for a period of time can be
calculated by finding the area bounded by the v=f(t) curve and the t-axis of
a v-t graph.

v
P

0a b c
Q t

v = f(t)

Total distance travelled from t = a to t = c
= Area P + lArea Ql



= න + න



Example 1

A particle is moving along a straight line. The displacement, s m, from a fixed
point O, at time t seconds, is given by s = 9t + t2.
Find
(a) the initial velocity of the particle,
(b) the velocity of the particle when it is 52 m to the right of fixed point O,

(a) Initial speed is the speed when t = 0.


= = 9 + 2
= 0, = 9 + 2 0 = −

(b) s = 52, 52 = 9t + t2
t2 + 9t - 52 = 0
(t - 4) (t + 13) = 0
t = -13 (rejected)
t=4

When t = 4, v = 9 + 2(4) = 17 m/s

Example 2

A particle is moving along a straight line. The displacement, s m, from a fixed
point O at time t seconds, is given by s = t3 - 6t2. Find
(a) the displacement when the particle stops for an instant,
(b) the range of values of t when the particle is moving toward the right.

(a) When the particle stops, v = 0.

= = 3 2 − 12


0 = 3 2 − 12

0 = 3 ( − 4)

= 0( ) = 4

ℎ = 4, = 43 − 6 42
= −

The answer shows that the particle is 32 m to the left of fixed point O.

(b) When the particle is moving toward the right, v is positive.

> 0, 3 2 − 12 > 0

3 ( − 4) > 0

04 t
t < 0 (ignored)
t > 4 seconds

Example 3

A particle moves along a straight line and passes through fixed point O. Its
velocity, v m/s, is given by v = 6t2 - 12t, at time t seconds after passing through
fixed point O. Find
(a) the displacement when the particle is at rest,
(b) the velocity when the particle passes through O again.

(a) = න

= න(6 2 − 12 )

= 2 3 − 6 2 +

t starts after the particle passes through point O  s = 0 when t = 0.
0 = 2(0)3−6(0)2+
= 0

, = 2 3 − 6 2

When the particle is at rest, v = 0.
0 = 6t2 - 12t
0 = 6t(t - 2)
t = 0 (ignored)
t=2

When t = 2, displacement, s = 2(23) - 6(22)
= -8 m

(b) When the particle passes through O again, s = 0.
0 = 2 3 − 6 2

0 = 2 2( − 3)

= 0 ( )
= 3

When t = 3, velocity, v = 6(32) - 12(3)
= 18 m/s

Example 4

A particle moves along a straight line from fixed point O. Its velocity, v m/s, is
given by v = 20 - 4t, at time t seconds after passing through point O. Find the
maximum displacement of the particle.

= න(20 − 4 )

= 20 − 2 2 +

t starts after the particle passes through point O  s = 0 when t = 0.

0 = 20(0) − 2(0)2+
= 0
, = 20 − 2 2

Displacement is maximum when velocity is zero.
v = 20 - 4t  0 = 20 - 4t

t=5
When t = 5, , = 20(5) − 2(5)2

=

Example 5

A particle moves along a straight line from a fixed point O. Its velocity, v m/s,
is given by v = 12 - 6t, at time t seconds after passing through O. Find the
distance travelled in the 4th second.

= න 12 − 6 = 12 − 3 2 +
s = 0 when t = 0, hence c = 0.
 s = 12t - 3t2

Distance travelled in the 4th second
= 4 − 3
= 12 × 4 − 3 × 42 − (12 × 3 − 3 × 32)

= −9

=

OR
s

s = 12t - 3t2
0 = 3t(4 - t)
t = 0 or 4 when s = 0

0 34 t

ℎ 4 ℎ

4

= න 12 − 6

3

= 12 − 3 2 4
3

= 12 × 4 − 3 × 42 − 12 × 3 − 3 × 32

= −9

=

Example 6

A particle is moving along a straight line. The displacement, s m, from the
fixed point O after t seconds is given by s = 2t - t2 + 8. Find the total distance
travelled in the first seven seconds.

First 7 seconds means from t = 0 to t = 7 seconds.

Find the time when the particle stops instantaneously (it stops in order to
reverse direction).


= = 2 − 2

= 0 = 2 − 2

= 1

When t = 0, s = 2(0) - (0)2 + 8 = 8 m
When t = 1, s = 2(1) - (1)2 + 8 = 9 m
When t = 7, s = 2(7) - (7)2 + 8 = -27 m

-27m O 8m 9m

Therefore, total distance in the first 7 seconds
= (9 - 8) + 9 + 27
= 37 m

OR s = - t2 + 2t + 8

s

9

8 7t
01

-27

Total distance in the first 7 seconds
= (9 - 8) + 9 + 27
= 37 m

Example 7

PQ

A 4m B

The diagram shows the positions of fixed points A and B on a straight line
where AB = 4 m. At one instant, particle P passes through point A with
velocity VP = 2t + 5, whereas particle Q passes through point B with velocity
VQ = 6 - t, such that t is the time in seconds after particles P and Q pass
through point A and point B respectively. Find

(a) the displacements of P and Q from fixed point A after t seconds,
(b) the distance between particles P and Q at the instant when Q stops,
(c) the time when both P and Q meet.

(a) For particle P : = න(2 + 5) = 2 + 5 +

= 0 ℎ = 0, = +

For particle Q : = න(6 − ) = 6 − 1 2 +
= 4 ℎ = 0, 2

= − +


(b) ℎ , = 0 ⇒ 6 − = 0

= 6

ℎ = 6, = 62 + 5 6 = 66
ℎ = 6,
= 6 6 1 62 + 4 = 22
−2

, = 66 − 22 =

(c) ℎ , =

2 + 5 = 6 − 1 2 + 4
2
3
2 2 − − 4 = 0

3 2 − 2 − 8 = 0

3 + 4 − 2 = 0

4 =
= − 3 ,

Example 8

A particle moves from rest from a fixed point O, along a straight line. Its
velocity, v m/s, is given by v = at2 - bt, where t is the time in seconds after
the motion starts, and a and b are constants. The particle rests
instantaneously when t = 2 second, and the distance of the particle at
that instant is 8 m to the left of O.
(a) Find the values of a and b.
(b) With the values of a and b found in (a),

(i) find the time taken by the particle to return to O,
(ii) sketch a displacement-time graph for 0 ≤ t ≤ 4,
(iii) calculate the total distance travelled in the first 4 seconds.

(a) = 0 ℎ = 2 ⇒ 0 = 22 − (2)
⇒ 0 = 2 − .........(1)

= න( 2 − )

3 2
= 3 − 2 +
3 2
= 0 ℎ = 0, = 3 − 2

= −8 ℎ = 2, (23) (22)
−8 = 3 − 2

⇒ −12 = 4 − 3 ............(2)

Solve (1) and (2) simultaneously, a = 6, b = 12

(b) (i) v = 6t2 - 12t
s = 2t3 - 6t2
When the particle returns to O, s = 0.
0 = 2t3 - 6t2
0 = 2t2(t - 3)

t=3s

(ii) s s = 2t3 - 6t2 0234
t 0 -8 0 32
32 s

0 2 34 t
-8

C - ACCELERATION

1 The acceleration of a particle moving along a straight line at a particular
instant is its instantaneous acceleration.

2 2
= = 2

3 (i) When a is positive, velocity increases with time.
(ii) When a is negative, velocity decreases with time.
(iii) When a is zero, the velocity is uniform.
Velocity is also maximum when acceleration is zero.

4 = න

2
5 2

= 0 ⇒ S = 0 ⇒ a

V
න න

ඵ

Example 1
A particle moves along a straight line and passes through fixed point O. The
velocity, v m/s, is given by v = t2 - 3t - 10, with time in seconds, t, after passing
the point O.
Find (a) the acceleration of the particle when its velocity is 8 m/s,

(b) the acceleration of the particle when the particle stops
instantaneously.

(a) 8 = 2 − 3 − 10
0 = 2 − 3 − 18
0 = ( − 6)( + 3)
= −3( )

= 6


= = 2 − 3

ℎ = 6, = 2(6) − 3

= −

(b) When the particle stops instantaneously, v = 0
0 = 2 − 3 − 10
0 = ( − 5)( + 2)
= −2( )

= 5

ℎ = 5, = 2(5) − 3

= −

Example 2

An object moves along a straight line and passes through the fixed point O. The

displacement, s m, is given by = 5 2 − 2 3 + 10 , where t is the time in
2 3

seconds after passing through point O. Find

(a) the time, t, when the velocity of the object is uniform,

(b) the range of time, t, when the object decelerates,

(a) When the velocity is uniform, acceleration, a = 0.

= 5 2 − 2 3 + 10 ,
2 3

= = 5 − 2 2 + 10



= = 5 − 4

0 = 5 − 4


=

(b) When object decelerates, acceleration, a < 0.

5 − 4 < 0


>

Example 3

The acceleration, a m/s2, of an object which moves along a straight line is given
by a = 8 - 2t, where t is the time in seconds after the object passes through fixed
point O. Given that the velocity of the object when passing through point O is
6 m/s, find the maximum velocity of the object.

= න = න 8 − 2
= 8 − 2 +
ℎ = 0, = 6 ⇒ = 6
, = 8 − 2 + 6

Velocity is maximum when a = 0.
0=8 - 2t
t=4

ℎ = 4, = 8(4) − (4)2+6

= 8(4) − (4)2+6
= −

Example 4

A particle moves along a straight line starting from fixed point O. The
acceleration, a m/s2, of the particle is given by a = 2t + 1, where t is the time in
seconds after the particle passes through point O. The initial velocity of the
particle is -6 m/s. Find the total distance travelled by the particle in the first 4
seconds.

= න = න 2 + 1

= 2 + +

ℎ = 0, = −6 ⇒ = −6
, = 2 + − 6

= න = න 2 + − 6

= 1 3 + 1 2 − 6 +
3 2

ℎ = 0, = 0 ⇒ = 0

, = 1 3 + 1 2 − 6
3 2

To find the total distance in the first 4 seconds, we find the displacement
when t = 0, t = 4 and the time when the particle reverses its direction (v = 0).

ℎ = 0, 0 = 2 + − 6

0 = ( − 2)( + 3)
= −3( )

= 2

= 1 3 + 1 2 − 6
3 2

1 2 3 + 1 (2)2−6(2)
(2) = 3 2

1
= −7 3

1 4 3 + 1 (4)2−6(4)
(4) = 3 2
t 024

1 s 0 1 1
= 53 −7 3 53

1 O1
−7 3 53

ℎ 4
111

= 73+73+53

=

The End