Question

a particle moving with uniform acceleration along a straight line ABC crosses point A at t=0 with a velocity 12mtps .B is 40m away from A and C is 64m away from A .The particle passes B at t=4s.After what time will the particle be at C?​

Answer 1

Step-by-step explanation:

Let the acceleration of the particle be a.

For motion between A and B

u= 12 m/s, s = 40 m, t = 4 s

s = ut + (1/2)at

2

⇒ 40 = 12 x 4 + (1/2) x a x (4)

2

a = -1 ms

−2

For motion between A and C

64= 12t + (1/2)(-l)t

2

=> r

2

- 24t +128 = 0

=>(t- 8) (t- 16) = 0

=> t = 8 s, 16 s

A.The particle will be at C twice, at t = 8 s and t = 16 s.