Question

A particle initially at rest moves with an acceleration of 5 m/s2. find the distance travelled in 5th second

Answer 1

u=0
a=5m/s²
t=5second
v=5
2as=v²-u²
2*5*s=5²-0²
10*s=25-0
10*s=25
s=25/10
s=2.5m

Answer 2

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