# Two car models are next to each other. First, the first model starts with acceleration a. After that, after time

**Two car models are next to each other. First, the first model starts with acceleration a. After that, after time t, the second model starts with an acceleration of 2a. At what distance S from the starting point will the models level up? Both car models travel in the same direction.**

Formula of uniformly accelerated motion for finding the path:

S = Vo * t + a * t ^ 2/2, where

Vo is the initial speed, a is the acceleration, t is the time.

Acceleration a and time T are set.

The initial speed of both machines is equal to zero according to the condition of the problem.

S = a * t ^ 2/2;

Path formula for the first car:

S1 = a * t1 ^ 2/2;

Path formula for the second car:

S2 = 2 * a * t2 ^ 2/2;

S2 = 2a * t2 ^ 2;

We’ll find out when S1 becomes equal to S2:

a * t1 ^ 2/2 = 2a * t2 ^ 2;

in this case, t1 = T + t2 (since the second car left at time T later).

(T + t2) ^ 2 = 4 * t2 ^ 2;

T ^ 2 + 2 * T * t2 + t2 ^ 2 – 4 * t2 ^ 2 = 0;

3 * t2 ^ 2 – 2 * T * t2 – T ^ 2 = 0;

Let’s solve the quadratic equation for t2:

D = 4 * T ^ 2 + 4 * 3 * T ^ 2;

D = 16 * T ^ 2;

First root of the equation:

t1 (1) = (2 * T + 4 * T) / (2 * 3);

t1 (1) = (6 * T) / (6);

t1 (1) = T;

Second root of the equation:

t1 (2) = (2 * T – 4 * T) / (2 * 3);

t1 (1) <0.

S = a * T ^ 2/2.

Answer: The distance at which the cars will align is a * T ^ 2/2.