Question

A body starts from rest and moves with uniform acceleration for 2 second . It then decelerates uniformly for 3 second and stops . If the deceleration is 4 m/s^2 the acceleration of body is?

Answer 1

Maximum velocity = v

During deceleration
v = u + at
0 = v - 4×3
v = 12m/s

During acceleration
v = u + at
12 = 0 + 2a
a = 6 m/s²

Acceleration of the body is 6m/s²

Answer 2

Answer:

correct answer is 6 ms^-2

Explanation:

Given final volume V = 0,

Deceleration a= 4 ms^-2

and

Deceleration time t= 3 s

Let U be the velocity before the deceleration is started

Using V= U+ at

0= U -4×3

U = 12 ms^-1

Now,

During acceleration, Initial velocity U1 =0,

Time = 2s,

Final volume U = 12 ms^-1,

Let acceleration is a

Using V= U + at

12= 0+a×2

a= 6 ms^-1

Hope it helps you my friend...........