Forces and Motion

Braking Distance of a Car - Velocity.

The braking distance of a car increases as the velocity increases.

Q1.  A car is moving with a velocity (speed) of 10 m/s.
When the brakes are applied,
the car has a constant negative acceleration (slows down) of -2 m/s2.
What is its stopping distance?

A1.  Use   a = (v-u) ÷ t
to see how long it takes the car to stop.

a = -2
v = 0
u = 10

t = (v-u) ÷ a
 t = (0 - 10) ÷ -2
 t = 10 ÷ 2

t = 5 seconds.

The car has constant acceleration,
so the average velocity = (initial velocity + final velocity) ÷ 2
= (10 + 0) ÷ 2
= 5 m/s.

Since velocity = distance ÷ time
distance = velocity x time
distance = 5 x 5
= 25 m.

Q2.  The same car is now moving with twice the velocity at 20 m/s.
When the brakes are applied,
the car has the same constant negative acceleration of -2 m/s2.
What is its stopping distance?

A2.  Use   a = (v-u) ÷ t

a = -2
v = 0
u = 20

t = (v-u) ÷ a
 t = (0 - 20) ÷ -2
 t = 20 ÷ 2

t = 10 seconds.

The average velocity = (initial velocity + final velocity) ÷ 2
= (20 + 0) ÷ 2
= 10 m/s.

Since velocity = distance ÷ time
distance = velocity x time
distance = 10 x 10
= 100 m.

Notice that doubling the velocity of the car from 10 to 20 m/s
has more than doubled the braking distance.
In fact the braking distance goes up x4 when the velocity goes up x2.
This is due to the car's kinetic energy.

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