Question

A car leaves Agra at 11 a.m. travelling at 50 km/h. how fast would a second car be travelling if it leaves Agra an hour later and overtakes the first car at 5 p.m. on the same day?

Answer 1

Hint: The speed and time of the first car are given, the distance it travels can be calculated by using the same. As the second car overtakes the first at the mentioned time, it will also travel the same distance as the first and from there, by applying the formula, its speed can be calculated.
Formula to be used:
 $ speed(s) = \dfrac{{dis\tan ce(d)}}{{time(t)}} $

Complete step-by-step answer:
First car leaves Agra at 11 a.m. and meets the second car at 5 p.m. , moving with a speed of 50 km/h.
Let the distance covered by both the cars from Agra till the meeting point be x, then according to the formula:
 $ \Rightarrow s = \dfrac{d}{t} $ and here,
Distance (d) = x
Time (t) = 6 hours (difference between 5 p.m. and 11 a.m.)
Speed (s) = 50 km/h
Substituting the values, we get:
 $ 50 = \dfrac{x}{6} $
x = (50 X 6) km
x = 300 km
Therefore, the distance travelled by both the cars from Agra to the meeting point is 300 km.
Now, the second leaves one hour after the first car,
The time of leaving of second hour = 11 a.m. + 1 hr
                                                                = 12 p.m.
The time of reaching the meeting point = 5 p.m.
The speed of this car can be calculated using the formula:
 $ \Rightarrow s = \dfrac{d}{t} $ and here,
Distance (d) = 300 km (calculated)
Time (t) = 5 hours (difference between 5 p.m. and 12 p.m.)
Speed (s) = S
Substituting the values, we get:
 $ \Rightarrow S = \left( {\dfrac{{300}}{5}} \right)km/h $
S = 60 km/h
Therefore, the second car will be travelling at a speed of 60 km/h, if it leaves Agra an hour later and overtakes the first car at 5 p.m. on the same day

Note: The units are always to be considered while solving the questions.
If the speed is in km/h, the distance calculated will be in km, provided the time is taken in terms of hours.
The time taken by the cars was the difference between leaving and reaching time as:
Net time = final time (reaching) – initial time (leaving)