Question

A body covers 12 m in the 2nd second and 20 m in the 4th second. Find what distance the body will cover in the 4 seconds after the 5th second.

Answer 1

sn= u + a/2 (2n -1)

s2 = 12

⇒ 12 = u+ a/2 (2 × 2 – 1)

⇒ 12 = u + 3 a/2 ………. (1)

s4 = 20

⇒ 20 = u + a/2 (2 × 4 – 1)

⇒ 20 = u + 7a/2 ………. (2)

Subtract equation (1) from (2)

• = 4a/2

⇒ a = 4 ms-2

12 = u + 3/2 × 4

⇒ u = 6 ms-1

Distance covered in the 4 seconds after 5th second,

= S9 – S5

= u (9) + 1/2 a (9)2 – [u(5) + 1/2 a (5)2]

= 4u + a/2 (81 – 25)

= 4 × 6 + 4/2 × 56 = 136 m.