COMEDK Exam 2018 Solved Paper

We know that, displacement travelled by a moving body with constant acceleration a in nth second,
sn=u+‌

1

2

a(2n−1)
For first second, n=1
∴‌‌s1‌=u+‌

1

2

a(2×1−1)
5‌=u+‌

a

2

...(i)[∵s1=5m]
For third second, n=3
∴‌‌s3=u+‌

a

2

(2×3−1)
2=u+‌

5a

2

. . . (ii) [∵s3=2m]
Subtracting Eq. (i) from Eq. (ii), we get
⇒‌‌‌−3=2a
‌⇒‌‌a=‌

−3

2

m∕ s2
Force acting on the body,
F=ma=4×(‌

−3

2

)=−6N
Hence, force acting on the body in opposite direction of its motion will be 6N.