Two bodies move according to the laws x1 = 3t + 0.5t ^ 2 and x2 = 12-6t. Determine the time and place of the meeting

x1 (t) = 3 * t + 0.5 * t ^ 2.

x2 (t) = 12 – 6 * t.

xbс -?

tbс -?

The bodies will meet at that moment in time when they have the same coordinates: x1 (t) = x2 (t).

3 * tbc + 0.5 * tbc ^ 2 = 12 – 6 * tbc.

0.5 * tbc ^ 2 + 9 * tbc – 12 = 0 – quadratic equation.

tbс ^ 2 + 18 * tbс – 24 = 0.

Let’s find the discriminant: D = 182 – 4 * (- 24) = 420.

tвс1,2 = (-18 ± √420) / 2.

tвс1 = 1.25 s, tвс ^ 2 = – 18 s – the root does not make sense.

hvs = 3 * 1.25 + 0.5 * (1.25) ^ 2 = 4.5 m.

xvs = 12 – 6 * 1.25 = 4.5 m.

Answer: the bodies will meet through tvs = 1.25 s, hvs = 4.5 m.