Who reaches the destination earlier? me or you?

Question

I drive a car from point A to B with the speed of 80 Km/h and return from B to A with the same speed. You drive the same route but your speed is 60 Km/h for A to B and 100 Km/h for B to A. Who will be in A again earlier? me or you? and why?

Answer 1

As Engman suggested above, suppose the distance from A to B is 100 km. You take 1.25 hours (75 minutes) to go each way, for 150 minutes total. I take 1.67 hours (100 minutes) to go and 1 hour (60 minutes) to return, for a total of 160 minutes. The first driver therefore makes the trip faster than the second driver.

The reason that you can't just average the speeds of the second driver is that although the second driver is driving the same distance at the two different speeds, he's not driving the same *time* at the two different speeds. You take average speed over time, not over distance.

If you drive 60 km/hour for one hour and then drive 100 km/hr for one hour, your average speed over the two hours will be 80 km/hr. But in this case the second driver is driving at the slower speed for more time than he's driving at the faster speed, so his average speed over his driving time won't be halfway between his slow speed and his fast speed; it will be less than the halfway point.

Suppose the distance from A to B is 80 km and the second driver drives for 1 hour at 60 km/hr and the next hour at 100 km/hr. In the first hour, he drives 60 km and in the second hour he drives 100 km, so after two hours he's traveled 160 km, the distance from A to B and back again. His average speed will be 80 km/hr and he will arrive at the same time as the first driver, whose constant speed was 80 km/hr.

Answer 2

It doesn't really matter what the distance is, but assuming a distance from A to B is one way to make the comparison and answer the question. I'm assuming the distance from A to B is 100 km.

YOU.
100 km x hr/80 km = 1.25 hr from A to B, same to return. So, your total time is 2.5 hours.

ME.
100 km x hr/60 km = 1.67 hr from A to B.
100 km x hr/100 km = 1.0 hr from A to B.
So, my total time is 2.67 hours.

ANSWER:
You will be at A again earlier than me, because it will take you less time.

Answer 3

Let distance from A to B be s km. ,then time taken to cover with speed 80 km /h will be s /80 hour.
The time taken to return from B to A will be same i.e. s /80 hour.
TOTAL time with speed 60 km/h= s/80 + s/80= s/ 40 hour.

In second case,
time taken to cover the distance s from A to B with speed 60 km /h will be s /60 hour.
the time taken to return from B to A will be s /100 hour.

In second case,TOTAL time will be= s/60 + s/100=2s/75 hour.
Thus time taken in second case is MORE THAN THE TIME TAKEN IN first case(with speed 80km/h fromA to B and back.
the first observer will return to A earlier than the second observer

Answer 4

This is a classic question in math texts and most students use intuition rather than physics and math. They invariably get the wrong answer.

When presented with such problems, an approach to provide an answer is to take an extreme example.

Suppose you traveled one way at 80km/hr and back at the same speed. Your average speed is 80km/hr. But now suppose you traveled at 160km/hr over the same distance. If you apply the same illogic that speeds may be averaged then you would travel at 0km/hr on the return trip and never get back.

Hope this helps to understand those who gave the correct answer that you get home first. Mike R

Answer 5

Well, Neither... We'll tie. The distance between A and B is fixed, so add the speeds. You go 80 and 80, I go 60 and 100. The average of my speed is 80 and the average of your speed is 80.

Answer 6

YOU WILL REACH EARLIER
LET IT BE 80 KM BETWEEN A,B
SO YOU CAN BE WITHIN 2HRS
I TAKE 80/60+80/100= 4/3+4/5 >2
HENCE PROVED

Answer 7

80x2= 160km/h
60+100= 160 km/h
The distance travled is the same and the total km/h is equal. Seems like a tie at the finish line.

Answer 8

Both arrive at the same time.

Answer 9

Don't need to do the math, you would get there first by 5 minutes.