A truck is passing by the stop in a straight street at a speed of 10. 5 seconds after the stop, a motorcyclist

A truck is passing by the stop in a straight street at a speed of 10. 5 seconds after the stop, a motorcyclist moving with constant acceleration goes after the stop and catches up with the truck after 150m. what is the acceleration of a motorcycle

Vg = 10 m / s.

t = 5 s.

V0m = 0 m / s.

S = 150 m.

a -?

Since the truck drives evenly, the distance that it traveled before meeting the motorcyclist can be expressed by the formula: S = Vg * tg, where Vg is the speed of the truck, tg is the time the truck travels before the meeting.

tg = S / Vg.

tg = 150 m / 10 m / s = 15 s.

Since the motorcyclist was moving uniformly accelerated, we will express the motorcyclist’s path to the meeting with the truck S by the formula: S = V0m * tm + a * tm2 / 2, where tm is the time of the motorcyclist’s movement before the meeting, and is the motorcyclist’s acceleration.

S = a * tm ^ 2/2.

a = 2 * S / tm ^ 2.

tm = tg + t.

tm = 15 s – 5 s = 10 s.

a = 2 * 150 m / (10 s) ^ 2 = 3 m / s2.

Answer: the acceleration of the motorcyclist is a = 3 m / s2.