Question

A car travelling at 60 km/h overtake another car travelling at 42 km/h. assuming each car to be 5 m long, find the time taken during the overtaking and the total distance used for the overtake.

Answer 1

Here both of the cars are travelling at constant speeds which does not change over the period of time. So, there is no acceleration involved and this is a case of motion with constant speed. Also, both the cars are moving and that too in the same direction so we have to use the concept of relative velocity in order to arrive at the solution.

Complete step by step answer:
Speed of first car, ${{v}_{1}}=60km/h$
$\Rightarrow {{v}_{1}}= 60\times \dfrac{5}{18} \\\ \Rightarrow{{v}_{1}}= \dfrac{50}{3}m/s \\\$
Speed of second car ${{v}_{2}}$= 42 km/h
$\Rightarrow {{v}_{2}}= 42\times \dfrac{5}{18} \\\ \Rightarrow {{v}_{2}}= \dfrac{35}{3}m/s \\\$
Now length of each car is 5m, so in order to pass one another completely the total distance to be covered is 5+5= 10 m
Now both are moving in the same direction, so the relative velocity, $v={{v}_{1}}-{{v}_{2}}$
v= \dfrac{50}{3}-\dfrac{35}{3} \\\ \Rightarrow v= \dfrac{15}{3} \\\ \Rightarrow v= 5m/s \\\
Now we can use the formula to find out the required time to do this,
t=\dfrac{d}{v} \\\ \Rightarrow t=\dfrac{10}{5} \\\ \therefore t=2s \\\
Now, the Distance travelled while overtaking= distance travelled in 2 seconds by car A + length of car B.
$\dfrac{50}{3}\times 2+5 \\\ \Rightarrow \dfrac{100}{3}+5 \\\ \Rightarrow 33.5+5 \\\ \therefore 38.33m$

So, the time taken to overtake the second car is 2s, and the total distance covered during this act is 38.33m.

Note: The speed was given in km/h while the length of the cars was m, so we convert the speed from km/h to m/s by multiplying with the conversion factor. Had the two cars been travelling in opposite directionsO then while finding out the relative velocity we would have added the two velocities instead of subtracting them as we had done in this problem.