Question

A motorbike is running at a velocity of 72kmh-1.On applying brakes, it is brought to rest in 2s.What is

the distance covered by the motorbike before coming to rest? Presume that acceleration produced by

brakes is uniform throughout.

10m

20m

5m

40m

​

Answer 1

Answer :

Initial velocity = 72km/h = 20m/s

Final velocity = zero

Time taken by bike to stop = 2s

We have to find distance covered by bike before it is brought to rest.

ATQ, Acceleration of bike is said to be constant so wr can apply equation of kinematics to solve this question.

First we have to find acceleration of bike$.$

» a = (v - u) / t

» a = (0 - 20) / 2

» a = -10m/s²

[፨ Negative sign shows retardation.]

Let's apply third equation of kinematics.

➺ v² - u² = 2as

➺ 0² - 20² = 2(-10)s

➺ -400 = -20s

➺ s = 20m

☃ God_Bless :)

Answer 2

Answer:

Initial velocity = 72km/h = 20m/s

Final velocity = zero

Time taken by bike to stop = 2s

We have to find distance covered by bike before it is brought to rest.

ATQ, Acceleration of bike is said to be constant so wr can apply equation of kinematics to solve this question.

First we have to find acceleration of bike..

» a = (v - u) / t

» a = (0 - 20) / 2

» a = -10m/s²

[፨ Negative sign shows retardation.]

Let's apply third equation of kinematics.

➺ v² - u² = 2as

➺ 0² - 20² = 2(-10)s

➺ -400 = -20s

➺ s = 20m

☃ God_Bless :)