By the end of the first second of uniformly slowed motion, the modulus of the body’s velocity is 2 m / s, and by the end

By the end of the first second of uniformly slowed motion, the modulus of the body’s velocity is 2 m / s, and by the end of the second – 1 m / s. Determine the modulus of the initial body velocity.

Given:

v1 = 2 meters per second – body speed by the end of the first second of equally slow motion;

v2 = 1 meter per second – the speed of the body by the end of the second second of equally slow motion.

It is required to determine v0 (meter per second) – the initial velocity of the body.

Let a be the acceleration of the body.

Then we have that:

v1 = v0 – a * t1, where t1 = 1 second.

v2 = v0 – a * t2, where t2 = 2 seconds.

From the first equation we find that: a = (v0 – v1) / t1. Then:

v2 = v0 – (v0 – v1) * t2 / t1;

v2 * t1 = v0 * t1 – v0 * t2 + v1 * t2;

v0 * t2 – v0 * t1 = v1 * t2 – v2 * t1;

v0 * (t2 – t1) = v1 * t2 – v2 * t1;

v0 = (v1 * t2 – v2 * t1) / (t2 – t1) = (2 * 2 – 1 * 1) / (2 – 1) =

= (4 – 1) / 1 = 3/1 = 3 meters per second.

Answer: the initial velocity of the body was 3 m / s.