Question

24. An object with an initial velocity V0
speeds up with an acceleration a, travelling a distance L1
, then it slows
down with a deceleration a, and stops after travelling an additional distance L2
. If L2 ÷L1=K,
then what is the
maximum velocity of the object during its travel?

Answer 1

Answer:

vmax=√(k/(k-1))vo.

Explanation:

From the question we get that the initial velocity of the body is given as vo and is speeding up in a acceleration of a. So, when the object is accelerating then the vmax^2=vo^2-2al1. For the retardation the initial velocity will be 0. Hence, the vmax^2 will be 2al2.

On dividing the two equations we will get that 2al2/2al1 = vmax^2/(vmax^2-vo^2). Since, l2/l1 is given in the question as k. So, we will get that k= vmax^2/(vmax^2-vo^2). Therefore, kvmax^2-kvo^2=vmax^2 or the vmax^2(k-1)=kvo^2. Which on solving we will get that the vmax=√(k/(k-1))vo.