Question

a particle starts moving from rest under a constant acceleration, if it travels a distance x in the sec. what distance will it travel in next t sec​

Answer 1

Answer:

Here initial velocity ,u=0

Distance, S=x

Acceleration, a=a

Time, t=t

According to second equation of motion

S=ut+12at2

⇒ x=0×t+12at2

Or x=12at2 −−−−(1)

Now when time is doubled

i.e. t=2t

Then \ S=12a(2t)2

Or S=4×12at2

Or S=4x [using value of x from equation (1)]

Thus distance covered in last t seconds= distance covered in 2t seconds - distance covered in first t seconds

Or S′=S−x=4x−x=3x

Another approach to this problem

Here initial velocity ,u=0

Distance, S=x

Acceleration, a=a

Time, t=t

According to second equation of motion

S=ut+12at2

⇒ x=0×t+12at2

Or x=12at2

Or at2=2x −−−−(1)

According to 1st equation of motion

Final velocity, v=u+at

Orv=0+at=at

Let Distance covered in next t seconds = S'

And initial velocity in this case is v

i.e. u=at

According to 2nd equation of motion

S′=ut+12at2

Or S′=at.t+12at2

Or S′=at2+12at2

Or S′=(1+12).at2

Or S′=32at2

Using value of at^2 from equation (1)

S′=32×2x=3x