A cyclist travels to work at an average speed of 3m/s and returns home for tea at an average speed of 9m/s. Calculate her average speed for the whole journey.(the answer is not 6m/s).
2007-03-02 19:00:10 · 7 answers · asked by jazmeen 2 in Science & Mathematics ➔ Physics
velocity is a vector meaning it has magnitude and direction.
in this case, it is measured in meters per second.
how many seconds did the journey take?
velocity= distance over time.
This is not a velocity question to start off with.
Nice try, at least-wouldnt total time be, final time minus initial time?
Honestly, there is no answer.
Alternately- you can almost say the answer is 4 m/s if you wanted to. 'Cause at work the cyclist isnt on a bike therefore v=0m/s, so 3+0+9 divide by 3 = 4m/s BUT no...none of these are correct! shame on those people who take other's work as their own. There is not enough information to answer this question. You should contact your teacher/lecturer on this one.
2007-03-02 19:06:26 · answer #1 · answered by Chiy9u 3 · 0⤊ 1⤋
If the distance to work is d, the time it takes to get to work is d/3. The time it takes to return id d/9. The total time is then d/3+d/9. In that time she has traveled a total distance of 2*d. Average velocity is total distance divided by total time, or 2*d/(d/3+d/9) = 2/(1/3+1/9); multiply both numerator and denominator by 9 to get 18/(3+1) = 18/4 = 4.5 m/s
2007-03-02 19:08:19 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋
As x procedures 2 from the beneficial or unfavorable component? It fairly does make a distinction. If from the unfavorable, the numerator would be an exceedingly small neg. style and denominator an exceedingly small neg. style and because x^3 will reason the numerator to attitude 0 swifter than (x-2) will, the decrease will attitude 0. From the beneficial, numerator will substitute right into a vey small beneficial and denominator an exceedingly small beneficial yet, the effect stands out as the comparable because of the fact the neg.. The decrease will returned attitude 0.
2016-10-17 04:02:48 · answer #3 · answered by balick 4 · 0⤊ 0⤋
Let distance from home to work be x m.
Time taken to work = x/3 s.
Time taken to home = x/9 s.
Total time taken = 4x/9 s.
Total distance = 2x m.
Therefore, average speed = total distance/total time =
2x divided by 4x/9 =
4.5 m/s.
2007-03-02 19:16:44 · answer #4 · answered by Xanana 3 · 0⤊ 0⤋
suppose the distance from the cyclist's home and workplace are x m apart,
t2(time taken for the cyclist to traverse the distance)=x/3
t3(for the cyclist to return home forom workplace to her home for tea)=x/9.
Total time taken =t=t2+t3
=x/3+x/9
=4x/9.
Avg. speed=dist /time
=2x/4x/9
=9x/2x=9/2m/s
=4.5 m/s
2007-03-02 19:10:45 · answer #5 · answered by Anonymous · 0⤊ 0⤋
well im no mathmetician but im going to say 6m/s even tho you said its not.
im just crazy like that.
2007-03-02 19:09:35 · answer #6 · answered by Anonymous · 0⤊ 2⤋
Do your own dang homework.
2007-03-02 19:35:01 · answer #7 · answered by smartprimate 3 · 0⤊ 0⤋