Question

14. A car accelerates uniformly from 18km/h to 36km/h in 2s. Calculate
a) The acceleration
b) The distance covered by the car in that time.​

Answer 1

acceleration = v-u/t

18km/h ------ 5m/s

36km/h------- 10m/s

10m/s-5m/s/2s

5/2

=2.5m/s^2

displacement of car = velocity/time

= 18/2 = 9

Answer 2

Given:- u=18km/h

=18×5/18

= 5m/s

v=36km/h

=36×5/18

=10m/s

t= 2sec

(a) To find acceleration

Using first equation of motion...

v=u+at

10=5+a×2

10=5+2a

10-5=2a

5=2a

5/2=a

a=2.5m/s^2

(b) To find the distance

Using second equation of motion.....

s=ut+1/2at^2

s=5×2+1/2×2.5×2^2

s=10+1/2×2.5×4

s=10+5

s=15m

Hope this answer will help you.....