Question

A 2 m long car is overtaking a 10 m long bus. The bus is travelling at a constant
speed of 20 m/s. The car takes 3 s to overtake the bus. We assume that
overtaking starts when the frontmost part of the car crosses the backmost part
of the bus and ends when the backmast part of the car crosses the frontmost
part of the bus. So at what constant speed (in m/s) Is the car driving?​

Answer 1

Given are the lengths of a car and bus, Find the speed of the car so that it can overtake the bus in three seconds for the given speed of the bus.

Explanation:

• Let there be an object moving with velocity 'v1' and another object moving with velocity 'v2'.
• Then the relative velocity of the first object to the second object is given by, $v_{12}=v_1-v_2$  
• Now, here since the direction of motion is the same for both the car and the bus.
• Therefore, we have their respective velocities as, $v_c,\ \ v_b=20\ m/s$.
• Hence the relative velocity of the car to the bus is, $V_{cb}=v_c-v_b$ .
• The overtaking ends when the backmost part of the car crosses the foremost part of the bus.
• Hence, the relative distance is given by, $d=10+2=12\ m$
• There for the time taken to overtake the bust by the car is, $t=\frac{d}{V_{cb}}=3\ s\\->V_{cb}=\frac{d}{t}\\ ->v_c-v_b=\frac{12}{3}\\ ->v_c=4+20\\->v_c=24\ m/s$  
• The constant speed at which the car is driving is $24\ m/s$.