Question

An object starts from rest and undergoes uniform acceleration. during the first second it travels 5.0 meters. how far does it travel during the third second

Answer 1

Distance traveled in third second = 25 meter.

Step-by-step explanation:

We have equation of motion,
`s= ut+(1)/(2) at^2`, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

During the first second it travels 5.0 meters

In this case u = 0 m/s, t = 1 second and s = 5 meters

Substituting

`5=0*1+(1)/(2) *a*1^2\\ \\ a=10m/s^2`

Now we need to find how much distance it travel during third second.

Distance traveled in third second = Distance covered in 3 seconds - Distance covered in 2 seconds

Distance covered in 3 seconds =
`0*3+(1)/(2) *10*3^2=45m`

Distance covered in 2 seconds =
`0*2+(1)/(2) *10*2^2=20m`

Distance traveled in third second = 45 - 20 = 25 meter