Question

9.
A car moving with
EAMCET & JEE MAINS VOL - TL
car moving with constant acceleration
covers the distance between two points 180
m apart in 6 sec. Its speed as it passes the
second point is 45 m/s. What is its acceleration
and its speed at the first point​

Answer 1

HERE IS YOUR ANS

let the initial velocity= u

v = 45m/s

by :- v = u + at

45 = u+6a

u = 45-6a-------(1)

by :- s = ut+1/2at²

by (1)

180 = (45-6a)6 + 1/2(a)36

180 = 270-18a

(a =5) m/sec²

so, initial velocity ,, U = 15m/sec.

hope it helps you ☺️