A motorist drives south at 30.0 m/s for 3.00 min, then turns west and travels at

Question

A motorist drives south at 30.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.20 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.20 min trip, find the following values. Let the positive x axis point east.

Find:

a) Total vector displacement and direction (in degrees) south of west.

b) Average speed.

c) Average velocity and direction (in degrees) south of west.

I kept getting an error message of "your response is within the 10% of the correct value" so I am feeling frustrated.

Explanation / Answer

distance travelled in
south = 5400
west = 3300
north-west = 1800

north component of northwest = 1800 x cos45 = 1272.79

total west component = 3300 + 1272.79 = 4572.79

total south component = 5400 - 1272.79 = 4127.21

total displacement = (4127.21^2 + 4572.79^2)^1/2 = 6159 m

angle = tan^-1 4127.21/4572.79 = 42.06 degree

avg speed=3*30*25*30/(30*25+25*30+30*30)=28.125

Avg velocity=6154(disp)/60*6.2(time)=16.5

angle = tan^-1 4127.21/4572.79 = 42.06 degree

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