In how many seconds from the beginning of the simultaneous movement of cyclists, the second

In how many seconds from the beginning of the simultaneous movement of cyclists, the second of them will catch up with the first, if their equations of motion are given by the dependencies X1 = 6 + 2 t and X2 = 0.5 * t ^ 2.

Given:

x1 (t) = 6 + 2 * t – the equation of motion of the first cyclist;

x2 (t) = 0.5 * t ^ 2 is the equation of motion for the second cyclist.

It is required to determine t (seconds) – the time interval after which the second cyclist will catch up with the first.

At the meeting point of two cyclists, their coordinates will be the same. Then:

x1 (t) = x2 (t);

6 + 2 * t = 0.5 * t ^ 2;

0.5 * t ^ 2 – 2 * t – 6 = 0;

t ^ 2 – 4 * t – 12 = 0 – we get a quadratic equation, where a = 1, b = -4 and c = -12.

Find the discriminant: D = b ^ 2 – 4 * a * c = (-4) ^ 2 + 4 * 1 * 12 = 16 + 48 = 64, D ^ 0.5 = 8

Find the roots of the equation:

t1 = (4 + 8) / 2 = 12/2 = 6, t2 = (4 – 8) / 2 = -2. (since t2 is less than zero, this root does not fit according to the problem statement).

Answer: the second cyclist will catch up with the first one 6 seconds after the start of the movement.