Question

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

2
. Particle
B starts 349 cm down the track and moves at
a constant velocity of 44 cm/s to the left.
When will they meet?
Answer in units of s.

Answer 1

t = 4 s

Step-by-step explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

`d = v_i t + (1)/(2)at^2`

`d = 0 + (1)/(2)(65.5)t^2`

`d_1 = 32.75 t^2 cm`

Now we know that B moves with constant speed so in the same time B will move to another distance

`d_2 = 44 * t`

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

`d_1 = 349 + d_2`

`32.75 t^2 = 349 + 44 t`

on solving above kinematics equation we have

`t = 4 s`