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

Question

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

Answer 1

Let the total distance covered at the end of t sec. be s[t].

Then:

s[t] = ut + at^2 / 2

The distance S[t] covered in the t-th sec. is s[t] - s[t - 1]:

S[t] = u[ t - (t - 1) ] + a[ t^2 -  (t - 1)^2 ] / 2

= u + a(2t - 1) / 2.

Putting S[2] = 12 and t = 2:

12 = u + 3a / 2 ...(1)

Putting S[4] = 20 and t = 4:

20 = u + 7a / 2 ...(2)

Subtracting (1) from (2):

8 = 2a

a = 4 m/s^2.

From (1):

u = 12 - 3a / 2

= 6 m/s^2.

In 5 sec the body has travelled distance x where:

x = 5u + a * 5^2 / 2

= 5u + 12.5a

In 9 sec the body has travelled distance y where:

y = 9u + a * 9^2 / 2

= 9u + 40.5a

In 4 sec. after 5 sec, the distance covered is:

y - x = 4u + 28a

= 24 + 112

= 136 m.