Question

A body covers a distance of 4m in 3rd second and 12 m in 5th second . if the motion is uniformly accelerated , how far will it travel in next 3 seconds ?

Answer 1

Let initial velocity be u and acceleration be a

sn = u + ½a(2n-1)

a/q

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.

Answer 2

sn = u + ½a(2n-1)

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