Question

A particle initially at rest, moves with an acceleration 5 ms–2 for 5s. Find the distance travelled in
(i) 4s
(ii) 5s
(iii) 5th second

Answer 1

Answer:

Explanation:

Using 1st equal of motion

vf=vi+at

vf=0+2 (5)

vf=10 m/s

Now use 3rd equation of motion

2aS=vf^2-vi^2

S=[vf^2-vi^2]/2a

S=[10^2-0^2]/2 (2)

S=100/4

S=25m