Two cars start uniformly accelerated movement from a state of rest in opposite directions with an acceleration

Two cars start uniformly accelerated movement from a state of rest in opposite directions with an acceleration of 0.2 m / s ^ 2. The second starts moving 2 seconds later than the first. At what speed is the second car moving in the reference system of the first one 5 s after the start of the first car?

Given:

a1 = a2 = 0.2 m / s2 – vehicle acceleration;

t = 2 seconds – the second car started moving 2 seconds later than the first;

t1 = 5 seconds – time span.

It is required to determine v (m / s) – the speed of the second car relative to the first.

Let’s find the speed of the first car:

v1 = a1 * t1 = 0.2 * 5 = 1 meter per second.

Let’s find the speed of the second car:

v2 = a2 * (t1 – t) = 0.2 * (5 – 2) = 0.2 * 3 = 0.6 meters per second.

Since, according to the condition of the problem, cars are moving in opposite directions, then:

v = v1 + v2 = 1 + 0.6 = 1.6 m / s.

Answer: the speed of the second car relative to the first is 1.6 m / s.