Question

A bus starts from rest with a constant acceleration of 5 m/sec Square. at the same time a car travelling at a constant velocity of 50m/sec overtakes the Bus and passes it. find at what distance will the bus overtake the car and how fast the bus would be travelling then

Answer 1

The law of the motion with the constant acceletation (from the origin of the coordinates and from rest):

$$x_{bus}(t) = \dfrac{at^2}{2}$$

where $x(t)$ is the bus coordinate at time $t$.

The law of the car's motion (constant speed, starts from the origin):

$$x_{car}(t) = vt$$

The time at which the bus will overtake the car satisfies the following consition:

$$x_{bus}(t_{overtake}) = x_{car}(t_{overtake})\\ \dfrac{at_{overtake}^2}{2} = vt_{overtake} \Rightarrow t_{overtake} = \dfrac{2v}{a} = \dfrac{2\cdot 50}{5} = 20 s$$

The distans of the meeting then:

$$d = x_{car}(t_{overtake}) = vt_{overtake} = 50\cdot 20 = 1000 m$$

The speed of the bus at that time will be:

$$v_{bus} = at_{overtake} = 5\cdot 20 = 100 m/s$$

Answer. d = 1000 m, v = 100 m/s.