Question

1
47) A body describes 0.12 m in
the 2nd second and 0.52 m in
the 4th second of its motion. If
the motion is uniformly
accelerated find how much
distance will it cover in 5th
second.​

Answer 1

Answer:

=u+a/2(2n−1)

s

2

=12

⇒12=u+a/2(2×2–1)

⇒12=u+3a/2 ………. (1)

s

4

=20

⇒20=u+a/2(2×4–1)

⇒20=u+7a/2 ………. (2)

Subtract equation (1) from (2)

s=4a/2

⇒a=4ms

−2

12=u+3/2×4

⇒u=6ms

−1

Distance covered in the 4 seconds after 5

th

second,

=S

9

–S

5

=u(9)+1/2a(9)

2

–[u(5)+1/2a(5)

2

]

=4u+a/2(81–25)

=4×6+4/2×56=136m.