Two bodies, the distance between which L = 60 m, begin to move simultaneously in one direction.

Two bodies, the distance between which L = 60 m, begin to move simultaneously in one direction. The first body moves with a speed V = 4 m / s, and the second, catching up with the first with an acceleration of a = 2 m / s * s. How long will it take for the second body to catch up with the first?

The laws of motion of bodies will look like this:

x = L + v * t – for the first body;

x = a * t ^ 2/2 – for the second.

The distance between the bodies is determined by the expression:

v * t – a * t ^ 2/2.

At the time when the second body catches up with the first distance between them will be equal to 0, we get the equation:

L + v * t – a * t ^ 2/2 = 0.

We substitute the known values:

t ^ 2 – 4t – 60 = 0.

t12 = (4 + – √ (16 – 4 * 1 * (-60)) / 2 * 1 = (4 + – 16) / 2;

t1 = -7 – has no physical meaning;

t2 = 10.

Answer: 10 s.