Average speed

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What is the average speed of a car moving from place "A" to place "B"
if it goes at a speed of 126 km / h in one direction and 70 km / h in the return direction?
a) 88 b) 90
c) 98 d) 100 ?
 
Average speed is the total distance divided by the total time for the trip.
 
What is the average speed of a car moving from place "A" to place "B"
if it goes at a speed of 126 km / h in one direction and 70 km / h in the return direction?
a) 88 b) 90
c) 98 d) 100 ?
Start with the equation

d(istance) = S(peed} * t(ime}

Assume that the distance between A and B = d km

time required to go from A to B = d/126 ....................................if you did not get stopped by a police officer.

time required to return from B to A = d/70

total time for trip (combined) = d/126 + d/70

total distance for trip (combined) = 2 * d

the average speed for trip (combined) = 2 * d / (d/126 + d/70)

continue........

Please show us what you have tried and exactly where you are stuck.​
Please follow the rules of posting in this forum, as enunciated at:​
Please share your work/thoughts about this assignment.​
 
What is the average speed of a car moving from place "A" to place "B"
if it goes at a speed of 126 km / h in one direction and 70 km / h in the return direction?
a) 88 b) 90
c) 98 d) 100 ?
I'm confused. If the starting point is A and the ending point is B, aren't you already told that the average speed from A to B is 126 km/h?

It appears that the problem is supposed to be asking, "What is the average speed of a car on a round trip between place A and place B". Did you quote the problem inaccurately? This trip is from A back to A, not to B.
 
Start with the equation

d(istance) = S(peed} * t(ime}

Assume that the distance between A and B = d km

time required to go from A to B = d/126 ....................................if you did not get stopped by a police officer.

time required to return from B to A = d/70

total time for trip (combined) = d/126 + d/70

total distance for trip (combined) = 2 * d

the average speed for trip (combined) = 2 * d / (d/126 + d/70)

continue........

Please show us what you have tried and exactly where you are stuck.​
Please follow the rules of posting in this forum, as enunciated at:​
Please share your work/thoughts about this assignment.​
I do expect some corner time from you for this response.
 
I assume that you really want the average speed going from A to B and then back to B.
We are not told the distance from A to B so call that "d" km. Since the speed from A to B is 126 km/h that will take \(\displaystyle \frac{d}{126}\) hours. Since the speed from B to A is 70 km/h that will take \(\displaystyle \frac{d}{70}\) hours.

So overall, you go a distance 2d in \(\displaystyle \frac{d}{126}+ \frac{d}{70}= \frac{5d}{630}+ \frac{9d}{630}= \frac{14d}{630}= \frac{d}{45}\) hours. Yes, that is an average speed of \(\displaystyle \frac{2d}{\frac{d}{45}}= 90\) km/h.
 
Corner time for you. You found the average speed to go from A to B and back to A. The problem is that the problem asked to find the average speed from A to B.
In my comment about that (#6), I said the problem was written wrong, but clearly was intended to ask for the round trip time.

Put the author in the corner.
 
Corner time for you. You found the average speed to go from A to B and back to A. The problem is that the problem asked to find the average speed from A to B.
As you know, "these" problems are generally translated (or re-interpreted) to English language.

The "the given" of the problem was "average speed of a car moving from place A to place B". Thus I applied the common sense extension of the "find" of the problem and went on......
 
When a problem is badly stated you can try a "common sense" change but you can't be certain it is correct.
 
Yes, my general advice (assuming that the original version of the problem was unclear) is to state your interpretation before showing your work, to make it clear what problem you are actually solving. (This is one of the reasons I dislike multiple-choice problems, or others with no place to show work or comments.)
 
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