Question

An object travelling 12 mls slows down at the
rate of 4 m/s². How long does it take to stop?​

Answer 1

Given:-

• Initial velocity ,u = 12m/s

• Final velocity ,v = 0m/s

• Retardation ,a = 4m/s²

To Find:-

• Time taken ,t

Solution:-

We have to calculate the time taken to stop the object.

Using 1st Equation of motion

• v = u +at

Where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Substitute the value we get

→ 0 = 12 + (-4) ×t

→ -12 = -4×t

→ t = -12/-4

→ t = 12/4

→ t = 3 s

Therefore, the time taken to stop the object is 3 second.

Extra Knowledge !!

Some equation of motion are

• v = u +at

• s = ut +1/2at².

• v² = u² + 2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement or Distance

t is the time taken

Answer 2

=> v = u +at

=> 0 = 12 + 4 × t

=> 12 = 4 ×t

=> 12/4 = t

=> 3 = t

=> t = 3s

Time taken is 3 second