Question

A body is moving with uniform velocity of 8 m/s when the body just crosses another body , the second one starts and moves with uniform accelertion of 4m/s2.distance coverd by the second body when they meet

Answer 1

Consider that the distance moved by the second body when it meets the first in the course of its motion is 'x'. Assume that we start the clock with time (t) right after the first body passes the second and the second begins accelerating. Our frame of reference is stationary with respect to the ground.


Now, for the second body, x = ut + (1/2)at²

Using the equation of motion, where u= initial speed of the body (0), 'a' its uniform acceleration and 't' the time after which they meet. 

Therefore, x=(1/2)(4)t²= 2t²


For the first body, the distance covered would be the same, since that's how they'll meet. The time 't' elapsed too remains the same.

For first body traveling with uniform velocity, t= x/8 sec.

(since speed= distance/time)

Therefore, considering the two equations,

x/8 = √(x/2)

or, (x²/16)=x/2

⇒2x²=16x
⇒x²=8x
⇒x(x-8)=0
⇒Either x=0 or x=8

Since x=0 implies the point when the bodies first meet (when the second just starts accelerating), x=8 m is the correct value.

Therefore, the bodies meet when the second body covers a distance of 8m from its position of rest at t=0 sec.

Answer 2

Answer:

Step-by-step explanation: