A car is travelling at constant velocity. Its brakes are then applied, causing uniform deceleration.

Question 31

A car is travelling at constant velocity. Its brakes are then applied, causing uniform deceleration.

Which graph shows the variation with distance s of the velocity v of the car?

Reference: Past Exam Paper – March 2016 Paper 12 Q7

Solution:

Answer: A.

The motion of the car can be broken down into two: one with constant speed and another with constant deceleration.

We need to identify the correct velocity-distance graph. So we need to find equations that relate the velocity and distance travelled.

Consider the car moving at constant velocity.

Speed / Velocity = distance / time

v = s / t

Since the velocity is constant, the line is obviously horizontal on a velocity-distance graph.

Now consider the car moving with constant deceleration.

Equation that relates the velocity to the distance:

v² = u² + 2as

Deceleration is negative acceleration, so the value of ‘a’ would be negative.

v² = u² – 2as

v = √(u² – 2as)

The line should satisfy the above relationship.

Plotting a graph of v against s gives a curve similar to that of option A.

For example, put u = 100 and a = 10 and plot points for x = 0 to x = 10. Joining the points would results in a curve as in option A. It has a negative gradient (due to the deceleration) with an increasing magnitude.

This can also be understood as such.

The gradient of a velocity-distance graph is

Gradient = Δv / Δs

We are told that the deceleration is uniform, i.e. the velocity changes by equal amount. So Δv is the same at any distance.

But what happens to the change in distance travelled (Δs) as the velocity decreases?

Consider the equation

s = ut + ½ at² 

Put the deceleration to be 10 m s^-2. That is, a = - 10 m s^-2

When initial speed u = 100 m s^-1, the distance travelled in 1 s is

s = 100(1) + (½ × -10 × 1²) = 95 m

When initial speed u = 50 m s^-1, the distance travelled in 1 s is

s = 50(1) + (½ × -10 × 1²) = 45 m

So, as the velocity decreases, the change in distance travelled each second becomes smaller.

Gradient = Δv / Δs

Since Δv is constant and Δs is decreasing with velocity, the value of gradient increases as the velocity decreases by equal amount.

Graph B shows the velocity-time graph of the car.

Graph C tells us that the distance travelled increases as velocity decreases.

Graph D indicates the car comes immediately to rest after travelling at constant speed.