Answer to Question #166173 in Physics for gibson mkandawire

Question

Just as a car start to accelerate from rest at a constant rate of 2.44m/s², a bus moving at a constant speed of 19.6m/s passes a car in a parallel lane.

(a) How long before the car overtakes the bus?

(b) How fast is the car going then?

(c) How far has the car gone at that point?

Answer 1

(a) The distance traveled by the car can be written as follows:

$$d_1=\dfrac{1}{2}at^2.$$

The distance traveled by the bus can be written as follows:

$$d_2=vt.$$

When the car overtakes the bus $d_1=d_2$ and we can write:

$$\dfrac{1}{2}at^2=vt,$$ $$t=\dfrac{2v}{a}=\dfrac{2\cdot19.6\ \dfrac{m}{s}}{2.44\ \dfrac{m}{s^2}}=16.1\ s.$$

(b) We can find the speed of the car when the bus overtaking from the kinematic equation:

$$v=v_0+at=0+2.44\ \dfrac{m}{s^2}\cdot16.1\ s=39.3\ \dfrac{m}{s}.$$

(c) We can find the distance that the car travels at that point from the kinematic equation:

$$d_1=\dfrac{1}{2}at^2,$$ $$d_1=\dfrac{1}{2}\cdot2.44\ \dfrac{m}{s^2}\cdot(16.1\ s)^2=316\ m.$$