Question

Initial velocity is 2.5m/sec with positive X direction and its accelerated uniformly at the rate 0.5m/s find distance in first 2sec,time to reach the ground if the velocity is 7.5m/sec ,distance it will cover in reaching if the velocity is 7.5m/sec?

Answer 1

Answer:

u=2.5m/s

a=0.5m/s^2

t=2sec

t=?

v=7.5m/a

s=?

By newtons 1st equation of motion

v=u+at

7.5=2.5+0.5×t

7.5-2.5=0.5×t

5=0.5×t

5÷0.5=t

10=t

therefore time required to reach the ground =10sec

By newtons 2nd equation of motion

s=ut+1÷2at^2

=2.5×2+1÷2×0.5×2×2

=5+1

=6m

therefore it covers a distance of 6m in 2 sec

PLEASE MARK ME AS BRAINLIEST