Question

A body moving with uniform acceleration covered 4m in 3rd sec and 12 min 5th sec.
find its acceleration.
Find the ratio between the distances travelled by the body in 5th second and 13th​

Answer 1

Answer:

Let initial velocity be u and acceleration be a

sn = u + ½a(2n-1)

according to question

u + ½a(2×3-1) = 4  ---1.

u + ½a(2×5-1) = 12  ---2.

Eqn. 2 – eqn. 1

=> ½a(2×5-1) - ½a(2×3-1) = 8

=> 4a = 16

=> a = 4 ms-2

By 1.

u + ½ ×4×5 = 4

=> u = -6 ms-1

Total distance covered 8 s

= -6×8 + ½×4×(8)2

= 80 m

Distance covered in 5th second

= -6×5 + ½×4×(5)2

= 20 m

So distance covered in next 3 second after 5th second will be

80 – 20 = 60 m

:) Hope this will help you.

Explanation: