Motion of a Car Under Uniform Acceleration
A car starts from rest and accelerates uniformly to a speed of 72 km/h over a distance of 500 m. Calculate the acceleration and the time taken to attain this speed.
If a further acceleration raises the speed to 90 km/h in 10 seconds, find this acceleration
and the further distance moved. The brakes are now applied to bring the car to rest under uniform
retardation in 5 seconds. Find the distance travelled during braking.
Solution
Given: u = 0, v = 72 km/h = 20 m/s, s = 500 m
Acceleration of the Car
Let a = acceleration of the car.
Using the equation:
v² = u² + 2as
(20)² = 0 + 2a × 500 = 1000a
a = (20)² / 1000 = 0.4 m/s²
Time Taken to Reach the Speed
Let t = time taken.
v = u + at
20 = 0 + 0.4 × t
t = 20 / 0.4 = 50 s
Further Acceleration from 72 km/h to 90 km/h
Given: u = 20 m/s, v = 96 km/h = 25 m/s, t = 10 s
Acceleration
v = u + at
25 = 20 + a × 10
a = (25 − 20) / 10 = 0.5 m/s²
Distance Moved
s = ut + 1/2 at²
s = 20 × 10 + 1/2 × 0.5 × (10)²
s = 200 + 25 = 225 m
Braking to Rest
Now consider the motion of the car when brakes are applied to bring it to rest in 5 seconds.
Given: u = 25 m/s, v = 0, t = 5 s
Distance Travelled During Braking
s = (u + v)/2 × t
s = (25 + 0)/2 × 5
s = 62.5 m
* The value 25 m/s used above is the final velocity from the previous case.