Question

a bus moving at a speed of 30ms^-1 is brought to rest within 10s by applying brakes. what is the retardation? Also calculate the distance travelled by bus before coming to rest?

Answer 1

Answers :-

Retardation of the body is 3 m/s²

Distance covered by the body is 150m

Given :-

• Initial Speed of the body is 30 m/s
• Time taken by the body is 10s
• As it applied brakes brought to rest { Final velocity is 0}

i.e

• u = 30
• v = 0
• t = 10s

To find :-

• Retardation of the body
• Distance covered by the body

Solution :-

As we need to calculate the retardation and distance of the body Firstly we calculate acceleration of the body by the first equation of motion that is

v = u+at

where ,

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time taken

Substituting the values ,

0 = 30+a(10)

10a +30 =0

10a = -30

a = -30/10

a = -3 m/s²

Acceleration of the body is -3m/s²  So,

Retardation of the body is 3 m/s²

Now finding the distance covered by the body from the 2nd equations of motion that is

s =ut+1/2at²

where ,

• s = distance covered
• u = initial speed
• t = time taken
• a = acceleration

Substituting the values ,

s = (30)(10) + 1/2 (-3)(10)²

s = 300 +1/2 (-3)(100)

s = 300 - 300/2

s = 300 -150

s = 150m

So, distance covered by the body is 150m