Question

A body covers 40 m in first second, 15 m in second and 35 m in third second. What is the average speed of the body ?

Answer 1

Explanation:

We know that distance Travelled by uniformly accelerated body in nths is given by

Sn=u+21a(2n−1),  where a is the acceleration of the body and u is the initial velocity 

it is given that Sn=4s at n=3

⟹4=u+25a  .............. eq(1)

Sn=12 at n=5s

⟹12=u+29a  ...............    eq(2)

solving eq(1) and eq(2),  we get

a=4m/s2 and  u=−6m/s

Now we have to find velocity at t=5s

we know that v=u+at

⟹v=−6+(4×5)=14m/s 

now distance travelled in next 3s can be calculated by 

S=ut+21at2         Note:    now u=14m/s

⟹S=14×3+214(32)=60m

Answer 2

Answer:

40+15+35/3=30

explanation:

40+15+35=90

90/3=30