Answer:
t = 4 s
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"
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]d = 0 + \frac{1}{2}(65.5)t^2[/tex]
[tex]d_1 = 32.75 t^2 cm[/tex]
Now we know that B moves with constant speed so in the same time B will move to another distance
[tex]d_2 = 44 \times t [/tex]
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
[tex]d_1 = 349 + d_2[/tex]
[tex]32.75 t^2 = 349 + 44 t[/tex]
on solving above kinematics equation we have
[tex]t = 4 s[/tex]
