Question

The velocity of the body startingfromrest is 40m/so in 2nd second and 12m/so in 10th second.what is distancetravelledby the body.what is the distancetravelled by the body.what is the distance travelledby the body between the 4th secondand 8th second

Answer 1

We know that distance Travelled by uniformly accelerated body in n

th

s is given by

S

n

=u+

2

1

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

it is given that S

n

=4s at n=3

⟹4=u+

2

5a

.............. eq(1)

S

n

=12 at n=5s

⟹12=u+

2

9a

............... eq(2)

solving eq(1) and eq(2), we get

a=4m/s

2

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+

2

1

at

2

Note: now u=14m/s

⟹S=14×3+

2

1

4(3

2

)=60m