CGPET 2019 Solved Paper

We know that, distance travelled in n th second
sn=u+

a

2

(2n−1)
Case 1 Given, n=3,s=12m
So,12=u+

a

2

(6−1)
⇒12=u+2.5a .......(i)
Case 2s=20m,n=5
So, 20=u+

a

2

(10−1)
⇒20=u+4.5a ........(ii)
from Eqs. (i) and (ii),
8=a(4.5−25)
⇒8=2a
⇒a=4m∕s2 .......(iii)
From Eqs. (i),
⇒12=u+25×4⇒u=2m∕s
From first equation of motion,
v=u+at
v=2+4×10=42m∕s