Question

A car is moving with a velocity of 45km\hr. The driver presses the brakes and the car comes to rest in 3sec. What is its retardation? How far does it move before coming to rest? Please do question

Answer 1

Concept:

• One dimensional motion
• Kinematics equation
• Retardation

Given:

• The initial velocity of the car u = 45 km/h = 45 * 1000/3600 = 45 * 5/18 = 12.5 m/s
• The final velocity of the car v = 0 m/s
• The time taken for the car to come to a rest = 3 s

Find:

• The retardation of the car

Solution:

We know the following kinematics equation

v = u + at

v = 0 m/s, u = 12.5 m/s, t = 3 s

0 = 12.5 + a (3)

-12.5 = 3a

a = -4.17 m/s^2

To find the distance it covers before coming to rest, we use the following equation.

v^2 = u^2 + 2as

0 = 12.5^2 + 2 (-4.17) s

8.33 s = 12.5^2

s = 18.75 m

The retardation is 4.17 m/s^2. The car covers a distance of 18.75 m before coming to rest.

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