# Velocity - time graph practice ```Velocity - time graph
8
6
4
Velocity m/s
1. The velocity – time
graph shows the motion
of a particle for one
minute. Calculate each of
the following.
2
0
30
-2 0
10
45
120
20
30
40
-4
50
45 60
70
-6
-8
Time (s)
(a) The acceleration in the first 10s.
(b) The total distance travelled.
(a) a = 6/10 = 0.6 ms-2.
(b) D = &frac12; 106+206+ &frac12; 156
+ &frac12; 156 = 240m
(c) The total displacement.
(c) s = 30 + 120 + 45 – 45 = 150m
(d) The average speed.
(d) = 240 60 = 4 ms-1
(e) The average velocity.
(e) = 150 60 = 2.5 ms-1.
Velocity - time graph
2. A car leaves a point O and accelerates uniformly from rest to a
speed of 6 ms-1 in 4 seconds. It then maintains a steady speed for
a further 10 seconds, after which it slows down to a halt in 2
seconds.
(a) Sketch a speed–time graph for the car’s journey.
Speed
ms-1
6
t=4
t=14 t=16 time (t)
Velocity - time graph
(b) Find the acceleration during the initial and final stages of
the journey.
Speed
ms-1
6
t=4
t=14 t=16 time (t)
The initial stage: Acceleration = gradient = 6  4 = 1.5 ms-2
The final stage: Acceleration = gradient = -6  2 = -3 ms-2
Velocity - time graph
(c) Find the total distance travelled by the car.
Speed
ms-1
6
12
t=4
60
6
t=14 t=16 time (t)
Distance travelled = Total area:
Area = &frac12; 6  4 + 6  10 + &frac12;  6  2 = 12 + 60 + 6 = 78 m
3. Two cars set off on a journey. The first car leaves at time t = 0,
where t is measured in seconds. It accelerates uniformly until it
reaches a speed of 8 m s–1 at t = 3 s. It then maintains a constant
velocity. The second car leaves from the same point at t = 3 and
travels with constant speed 16 m s–1.
(a) On the same axes sketch speed–time graphs for the motion of the
two cars.
speed
(m s–1)
16
second car
8
first car
3
time (t)
(seconds)
(b) Find t when the two cars meet.
speed
(m s–1)
16
second car
8
first car
3
time (t)
(seconds)
Suppose the cars meet after t seconds:
First car: s = &frac12;  8 3+ 8(t – 3)
Second car: s = 16(t – 3)
16(t – 3) = &frac12;  8 3+ 8(t – 3)  16t – 48 = 12 + 8t - 24
8t = 36  t = 4.5 s
(c) How far are they from the start at this time ?
Distance travelled by the first car = 16( t – 3)
= 16( 4.5 – 3) = 24m
So, they meet 24 m from the start.
d) Sketch a displacement–time graph for the journey of the
second car.
Displacement m
0
3
time (t)
4. The graph shows the motion of a bungee jumper.
30
Velocity m/s
20
10
50
0
0
1
2
3
4
85
6
-10
Time (s)
(a) State two non-zero times at
which the velocity of the jumper
is zero.
(a) t = 4 s and t = 5.6 s
(b) Find the total distance
travelled and the final (b) D = &frac12; 4 25 + &frac12; 1.610 = 58m
displacement.
s = &frac12; 4 25 - &frac12; 1.610 = 42m
below starting position.
```