Question

A body has started from rest and attains an acceleration of 3 m/s2. find the total distance traversed if the final velocity is 6 m/s.

Answer 1

Answer:

Given that,

Acceleration a=2m/s

2

Retardation a=−3m/s

2

Time t=10s

Now, from equation of motion

The maximum velocity attained be v at t

v=u+at

v=0+2×t

v=2tm/s.....(I)

Now, again it reaches velocity 0 after 10 s with retardation

v=u+at

0=v−3(10−t)

Now from equation (I)

2t−3(10−t)=0

2t−30+3t=0

5t=30

t=6s

Now, put the value of t in equation (I)

v=2t

v=2×6

v=12m/s

Hence, the maximum speed is 12 m/s.. pls mark as brainlist