Question

A body starts from rest and moves for 5 seconds. It accelerates uniformly at the rate 2.5m/s2. Find its average velocity

Answer 1

Avg speed × total times = total distance

20×25=500

Since acceleration in the 1st interval is same as deceleration in the 3rd interval and deceleration is from the same velocity achieved after acceleration, t

1

=t

3

s

1

=

2

1

×5t

1

2

s

2

=(5t

1

)t

2

s

3

=(5t

1

)×t

3

−

2

1

×5×t

1

2

=(5t

1

)×t

1

−

2

1

×5×t

1

2

s

1

+s

2

+s

3

=5t

1

t

2

+5t

1

2

⇒5t

1

2

+5t

1

t

2

=500−−−−1

2t

1

+t

2

=25−−−−2

Solving (1 ) and (2) ⇒t

1

=5,t

2

=15sec

Answer 2

Answer:

6.25 m/s

Explanation:

apply v=u+at

v= ?

u=0

a=2.5 m/s²

t= 5 sec

from v=u+at , we get v=12.5m/s

now average velocity= v+u/2

12.5+0/2= 6.25m/s