The initial velocity of the body is 2m / s, its acceleration is 0.5m / s ^ 2. How many times will the

The initial velocity of the body is 2m / s, its acceleration is 0.5m / s ^ 2. How many times will the body’s speed increase on the 32m path?

V0 = 2 m / s.

a = 0.5 m / s2.

S = 32 m.

V / V0 -?

With uniformly accelerated motion, the body’s acceleration a shows how quickly the body’s velocity changes over time t.

The acceleration of the body a can be expressed by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * S, where V, V0 are the final and initial speed of movement, S is the path traveled by the body.

V ^ 2 – V0 ^ 2 = 2 * S * a.

V ^ 2 = V0 ^ 2 + 2 * S * a.

V = √ (V0 ^ 2 + 2 * S * a).

V = √ ((2 m / s) ^ 2 + 2 * 32 m * 0.5 m / s2) = 6 m / s.

V / V0 = 6 m / s / 2 m / s = 3.

Answer: the speed of the body has increased 3 times: V / V0 = 3.