Question

At a particular instant, a particle moving
with
a constant velocity is approaching a fixed point with
a velocity u-3m/s and after time ∆t=6s the
Particle passes position closest to fixed pt with
velocity u=5m/s find distance between. fixed point and particle
present? ​

Answer 1

It has given that,

initial velocity, u = 3 m/s

final velocity, v = 5 m/s

time taknen, ∆t = 6s

To find : The distance between fixed point and present point.

solution : first we have to find the acceleration of particle.

using formula, v = u + at

here, u = 3 m/s , v = 5 m/s and t = 6s

so, 5 m/s = 3 m/s + a × 6 sec

⇒2 = 6a

⇒a = 1/3 m/s²

now distance between fixed point and present point, s = ut + 1/2 at²

⇒s = 3 m/s × 6s + 1/2 × 1/3 m/s² × (6 s)²

⇒s = 18 m + 6 m = 24 m

Therefore the distance between fixed point and present point is 24 m