A truck is passing by the stop on a straight street at a speed of 10 m / s. After 5 seconds from the stop

A truck is passing by the stop on a straight street at a speed of 10 m / s. After 5 seconds from the stop, a motorcyclist moving with an acceleration of 3 m / s2 drives off after the truck. At what distance from the stop will the motorcyclist catch up with the truck?

Given:

v = 10 meters per second – truck speed;

t1 = 5 seconds – the time interval after which a motorcyclist drives after the truck;

a = 3 m / s2 – the acceleration of the motorcyclist.

It is required to determine L (meter) – at what distance from the stop the motorcyclist will catch up with the truck.

The truck motion equation will look like:

x = v * t1 + v * t = 10 * 5 + 10 * t = 50 + 10 * t, where v * t1 is the distance that the truck will have time to cover in time t1.

The equation of motion for a motorcyclist will be:

x = a * t ^ 2/2 = 3 * t ^ 2/2.

Let’s find the time after which the motorcyclist will catch up with the truck:

50 + 10 * t = 3 * t ^ 2/2;

100 + 20 * t = 3 * t ^ 2;

3 * t ^ 2 – 20 * t – 100 = 0;

D = 20 ^ 2 + 4 * 3 * 100 = 400 + 1200 = 1600; D0.5 = 40.

t1 = (20 + 40) / 6 = 60/60 = 10 seconds.

t2 = (20 – 40) / 6 = -20 / 6 – does not fit according to the problem statement.

Then the path will be equal to:

L = a * t1 ^ 2/2 = 3 * 10 ^ 2/2 = 3 * 100/2 = 3 * 50 = 150 meters.

Answer: The motorcyclist will catch up with the truck at a distance of 150 meters from the stop.