Question

A body covers 12m in 2nd second and 20 m in 4th swcond . How much distance will it covers in 4 second after 5th second????

Answer 1

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As we have a formula of motion that distance covered in t th second is

s=u+a(2t−1)/2" role="presentation" style="margin: 0px; padding: 0px; outline: 0px; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">s=u+a(2t−1)/2s=u+a(2t−1)/2

12 = u + a(2*2 - 1)/2
=> 12 = u + 3a/2
=> 2u + 3a = 24

and

20 = u + a(2*4 - 1)/2
20 = u + 7a/2
=> 2u + 7a = 40

From 1 and 2:
a = 4
u = 6

Now, distance covered in 4s after 5th second = Distance covered in 9 seconds - distance covered in 5 seconds

= 6*9 + 1/2 * 4 * 9 * 9 - (6 * 5 + 1/2 * 4 * 5 * 5)
= 54 + 162 - (30 + 50)
= 136 m



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