Question

a body covers a distance of 50 m in its 4th sec starting from U with same initial velocity and acceleration covers a distance of 80 m in its 8th sec . find the distance convered by the body in 3rd sec

Answer 1

s=ut + 1/2at²

s= 4u + 1/2×a×4²

50= 4u+ 8a

u + 2a = 50/4

2u + 4a = 25-------------------------1

Again,

s= ut + 1/2at²

s= 8u + 1/2×a×8²

80 = 8u + 32a

u+ 4a = 80/8

u + 4a = 10----------------------------------2

eq.2 - eq.1

(2u+ 4a) - (u + 4a) = 25 - 10

u = 15 m/s

u + 4a = 10

15+ 4a = 10

4a = -5

a = -5/4 m/s²

Again ,

s = ut+ 1/2at²

s = 15×3 + 1/2× -5/4 ×9

s = 45 - 45/8

s = 315/8

s = 39.375 m