A man walks 440 m

Throwback24 · Dec 3, 2008

1. The problem statement, all variables and given/known data

A man walked 440 m 50 degrees polar positive, and then 580 m 185 degrees polar positive. The entire trip took 150 min.

Total distance travelled?
The displacement of the man?
Average of the man in m/min?
Average velocity of the man in m/min?

2. Relevant equations

Speed=Total Distance/Total Time
Average velocity=Displacement/Total Time

3. The attempt at a solution

I graphed this out. I was accustomed to distance being N/S/W/E. Distance is a scalar quantity so it doesn't really into account positive and negative.

Would my total distance travelled be 1020 m?

Help please?

c. 1020 m/150 mins=6.8 m/s

Carid · Dec 3, 2008

a. Correct. The man walked 440m then 580m so he walked a total of 1020m.
b. However at the end of his walk he is not 1020m from his starting point. So how far is he?
c. You are correct.
d. Velocity is a vector quantity. We need the sum of his two velocity vectors. Or more simply the displacement from question b divided by the total time.

Throwback24 · Dec 3, 2008

I'm confused as how to calculate displacement for this, can you give me an example please?

mmmboh · Dec 3, 2008

C would actually be 6.8m/min...just a small mistake but you might lose marks :P

mmmboh · Dec 3, 2008

To calculate the displacement you can make a triangle and have the hypotenus be his distance, then you can do 440 sin 50 degrees as his first displacement, and then I believe you would add to that 580 sin 185 degrees (which would be a negative number)...I believe his total displacement would be 286.5 m....I'm not positive though.

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A man walks 440 m

Throwback24

Carid

Throwback24

mmmboh

mmmboh

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