Question

from the
27.
A body leaving a certain point O moves with
acceleration, which is constant in magnitude and direction.
At the end of the fifth second, its velocity is 1.5 ms-1. At the
end of the sixth second, the body stops and then begins to
move backwards.
(a) Find the distance traversed by the body, before it stops.
(b) Determine the velocity with which the body returns to
[Ans.(a) 27 m; (b) 9 m s-11
hundred metre sprinter increases her speed from rest
three quarters of
point O.​

Answer 1

Answer:

velocity = -9m/s

distance = 27m

Explanation:

The velocity slows 1.5 m/s in the sixth second so if the initial velocity direction is positive, the acceleration is

 a = (dv/dt) = -1.5 m/s²

Using the equation, v = u + at and knowing that the return time to zero velocity will equal the approach time of 6 seconds.

v = 0 + (- 1.5) × 6 v = -9 m/s

So the velocity upon returning to "O" is -9 m/s.

The average velocity between "O" and the stop point is v = (9 + 0) / 2 v = 4.5 m/s

So the particle travels d = 4.5 × 6 m → d = 27 m between "O" and the stop points