A bus driver heads south with a steady speed of v1-22.0 m/s for ti-3.00 m northw

Question

A bus driver heads south with a steady speed of v1-22.0 m/s for ti-3.00 m northwest at v3-30.0 m/s for t3-1.00 min. For this 6.20-min trip, calculate the following. Assume +x is in the eastward direction. then makes a ight turn and tre els at v2-2SO m/s for ta-2.20 mm s d then an es total vector displacement (Enter the magnitude in m and the direction in degrees south of west.) magnitude direction (a) ° south of west (b) average speed (in m/s) m/s (c) average velocity (Enter the magnitude in m/s and the direction in degrees south of west. magnitude direction m/'s o south of west

Explanation / Answer

Distance= speed* time

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d1= 22*3*60 (-j)= - 3960 j

d2= 25*2.2*60(-i)= -3300 i

d3= 30*60*( - cos 45 i + sin 45j)= -1272.8 i + 1272.8j

net displacement,

d= d1+d2+d3= - 4572.8 i - 2687.2j

Magnitude,

d= 5304 m. (Ans)

Direction, x= arctan(2687.2/4572.8)= 30.44 south of west.

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Av speed= total distance travelled/ total time taken= (3960+3300+1800)/(6.2*60)= 24.355 m/s

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Av velocity= total displacement/time= 5304/(6.2"60)= 14.26 m/s

Angle= 30.44 degrees

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Comment in case any doubt.. good luck

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