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At 10.30 a.m, a van left Town X travelling at an average speed of 64 Km/h.

At 11.15 a.m., a car left Town X, travelling on the same road at an average speed of
80 Km/h.
a) At what time did the car catch up with the van?

b) How far from Town X did each vehicle travel when they passed each other?

User Jdepypere
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1 Answer

4 votes

Answer:

a) 2:15 pm

b) 240 km

Explanation:

You want to know the time and place where a car leaving at 11:15 a.m. at 80 km/h catches up with a van leaving at 10:30 a.m. at 64 km/h.

Head start

The van travels for 11:15 -10:30 = :45, or 3/4 hour, before the car starts. This gives it a distance advantage of (3/4 h)(64 km/h) = 48 km.

Closing speed

The speed at which that distance is reduced is the difference between the car speed and the van speed:

80 km/h -64 km/h = 16 km/h

Closing time

The time it takes for the head-start distance to be reduced to zero is ...

time = distance/speed

time = (48 km)/(16 km/h) = 3 h

a) Meeting time

Three hours after the car leaves, it will catch up with the van. That time is ... 11:15 +3:00 = 14:15 = 2:15 p.m.

b) Meeting distance

In 3 hours, the car travels (3 h)(80 km/h) = 240 km.

Note that the van has been traveling 3 3/4 hours, so will have also traveled (3 3/4 h)(64 km/h) = 240 km. The two vehicles need to be in the same place at the same time if they are to pass each other.

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Additional comment

The attached graph shows the two vehicles will have traveled 240 km when they mean at 2:15 pm. The horizontal axis is hours after midnight. The vertical axis is kilometers from town X. The relation graphed is distance = speed × time.

At 10.30 a.m, a van left Town X travelling at an average speed of 64 Km/h. At 11.15 a-example-1
User Bryan Norden
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