Question

a body starting from rest starts running and attains a velocity of 6m s^-1 in 30 second . Then he slows down uniformly to 4m s^1 in next 5 s calculate his acceleration in both cases.​

Answer 1

Answer:

The acceleration in the first case is 0.2 m/s² and in the second case is -0.4 m/s²

Explanation:

Given :

• A body starting from rest starts running and attains a velocity of 6 m/s in 30 seconds.

• Then he slows down uniformly to 4 m/s in next 5 s

To find :

the acceleration in both cases

Solution :

The acceleration is the rate of change of velocity.

 $\boxed{\longrightarrow \tt a=\dfrac{v-u}{t}}$

where

v denotes the final velocity

u denotes the initial velocity

t denotes the time

In the first case,

The body started from rest and attained velocity of 6 m/s in 30 seconds.

u = 0 m/s

v = 6 m/s

t = 30 sec

Acceleration is : (let it be a₁)

a₁ = (6 - 0)/30

a₁ = 6/30

a₁ = 1/5

a₁ = 0.2 m/s²

In the second case,

The body slows down to 4 m/s in next 5 seconds.

u = 6 m/s

v = 4 m/s

t = 5 sec

Acceleration is : (let it be a₂)

a₂ = (4 - 6)/5

a₂ = -2/5

a₂ = -0.4 m/s²

Therefore, the acceleration in the first case is 0.2 m/s² and in second case is -0.4 m/s²