Moving with an acceleration of 0.5 m / s2, the body on the path of 60 m increased its speed

Moving with an acceleration of 0.5 m / s2, the body on the path of 60 m increased its speed 4 times. Find the initial velocity of the body.

a = 0.5 m / s2.

V = 4 * V0.

S = 60 m.

V0 -?

When the body moves uniformly accelerated, then its traversed path is determined by the formula: S = (V^2 – V0^2) / 2 * a, where V0, V are the initial and final velocity of the body, a is the acceleration of the body during motion.

Since by the condition of the problem V = 4 * V0, the formula will take the form: S = ((4 * V0) ^2 – V0^2) / 2 * a = (16 * V0^2 – V0^2) / 2 * a = 15 * V0^2 / 2 * but.

V0^2 = 2 * a * S / 15.

We find the initial speed of the body by the formula: V0 = √ (2 * a * S / 15).

V0 = √ (2 * 0.5 m / s2 * 60 m / 15) = 2 m / s.

Answer: the initial speed of movement was V0 = 2 m / s.