Uniform Motion Related Problem

A car travelling at 60 kph left town at 9:00 a.m. Another car travelling at 65 kph left the same place half an hour later to overtake the first. At what time did the second car overtake the first?

d1 = r1*t1 = 60km/h * 0.5h = 30 km after 0.5 h.

r2*t = r1*t + 30
65 * t = 60*t + 30
65t - 60t = 30
5t = 30
t = 6 hours.

To solve this uniform motion related problem, we can set up equations representing the distances traveled by each car.

Let's assume that the second car overtakes the first car in t hours after it leaves town.

The first car travels at a speed of 60 kph for t hours and has a head start of 0.5 hours. So, the distance traveled by the first car is D1 = 60(t + 0.5) kilometers.

The second car travels at a speed of 65 kph for t hours. Therefore, the distance traveled by the second car is D2 = 65t kilometers.

Since the second car overtakes the first car, the distance traveled by the first car and the distance traveled by the second car must be equal.

Setting up the equation:
D1 = D2

60(t + 0.5) = 65t

Now, let's solve the equation to find the value of t:

60t + 30 = 65t
30 = 5t
t = 6

Therefore, the second car overtakes the first car 6 hours after it left town.

To find the time at which the second car overtakes the first car, we need to add this time to the time the first car left town.

The first car left town at 9:00 a.m. and took 6 hours to be overtaken, so the second car overtakes the first car at 3:00 p.m.