Question

a man drives Northward at the rate of 80 km/h and then back to his starting point at the rate of 60 km/hr. What is his average velocity?

Answer 1

Hello @celyn !
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Answer 2

\(\large \sf Average~Velocity=\frac{Total~Displacement}{Time~Taken}\\Here,~Total~displacement=0~because~man~returns~to~his~starting~point\\=>Average~velocity=\frac{0}{Time~taken}=0~m/s\)

Answer 3

Let the distance to the furthest point traveled northward be d.
The time t1 taken when going at 80 km/h is given by:
\[\large t _{1}=\frac{d}{80}\]
The time t2 taken when going at 60 km/h is given by:
\[\large t _{2}=\frac{d}{60}\]
Total time for the round trip is:
\[\large t _{1}+t _{2}=\frac{d}{80}+\frac{d}{60}=\frac{7d}{240}\]
The total distance traveled on the round trip is 2d. Therefore the average velocity is given by:
\[\large \frac{total\ distance}{total\ time}=\frac{2d}{\frac{7d}{240}}=\frac{2d \times240}{7d}=68.57\ km/h\]

Answer 4

kropot72 I'm still confused. You didn't include the directions

Answer 5

@kropot72

Answer 6

It doesn't matter which direction he drove in- the question says he drove back to his starting point. You have to assume he drove straight there and back.

Answer 7

so therefore... the average velocity is zero?

Answer 8

I think the answer required is that given by kropot. i.e. based on total distance travelled, rather than total displacement.

However it IS true that the total change of position in the time is 0, but I believe this is a slightly pedantic interpretation of the question. (Though strictly correct)

If this is part of a test I would go for the total distance covered/total time answer.

If part of a class - discuss it with your tutor

Answer 9

@celyn
The question is perfectly ok.....The average velocity will be 0. Average speed will be what @kropot72 calculated.

\(\huge \bigstar \sf Remember\bigstar \\ \large Average~Speed=\frac{Total~Distance}{Time~Taken}\\ ~ \\~ \\ \large Average~Velocity=\color{blue}{\frac{Total~Displacement}{Time~Taken}}\)

Answer 10

\(\color{blue}{\text{Originally Posted by}}\) @MrNood
It doesn't matter which direction he drove in- the question says he drove back to his starting point. You have to assume he drove straight there and back.
\(\color{blue}{\text{End of Quote}}\)


It doesn't matters if he drove straight or took a curved path and there is no need to assume anything. The point is he drove back to the same position from where he started which means his displacement is 0

Answer 11

if he returned to his initial position then = 0.

Answer 12

Displacement is the straight line distance between the initial position and final position.. so if initial and final positions are the same, then Displacement = 0.

Answer 13

The average velocity is indeed zero, the reason being that the total displacement is zero.
\[\large \bar{v}=\frac{d}{t}=\frac{displacement\ (a\ vector)}{elapsed\ time\ (a\ scalar)}\]

Answer 14

yes and 68.57 is the `average speed`

Answer 15

IF this is an online test and the re is one chance to put an answer - my guess is that the intention of the question is that the OP works out the total distance/time .

i.e. I would expect that the 'correct answer as expected by the test is 68.6 kph

Sometimes one has to interpret the question - not all questions are precise.

Whatever your stance here - if I were taking an online test with one chance - I would NOT choose the 0 answer.

Answer 16

@MrNood That is the basis for my first post. I was well aware of the precise answer, but considered the answer of zero to be too easy.

Answer 17

Thanks for the clarifications guys! I'm just really confused :)

Answer 18

@Celyn
Yes - I can see that this is an inconclusive from your point of view.


Maybe you could put the question in perspective -
Is it an online multichoice test?
Is it classwork to be marked by tutor?

What are you currently studying? speed/distance/ time releationships ? or maybe difference between speed and velocity? or maybe difference between distance covered and displacement?

What do the other questions that you are currently given in the same class?

Answer 19

I'm not sure what's wrong but our teacher marked my answer wrong. I did what you guys just suggested, and I believe that the average velocity in this problem is 0. She told me that the formula for average velocity is (lol sorry I can't type it so I just draw it..hehe) She said the answer was this : :(((

Answer 20

I'm afraid that your teacher has it wrong.

Suppose you drove on a straight road for 1 hour at 60 mph and then for 5 hours at 40mph.

you will cover 60 miles+ 5*40miles = 260 miles

The time taken is 6 hours - so the average velocity is 260/6 = 43.33mph

If you used her equation it would be 50mph - but oyu have been travelling for 6 hours and have NOT covered 300miles therefore your average is NOT 50.

The reason she is wrong is because the time is INVERSELY related to speed.