A chauffeur heads south with a steady speed of v - 24.0 m\'s for t 3.00 min, the

Question

A chauffeur heads south with a steady speed of v - 24.0 m's for t 3.00 min, then makes a right turn and travels at for t) " 1.00 min. For this 7.00-min trip, calculate the folowing. Assume .zis in the eastward direction. -25.0 mys for t3.00 min, and then drives northwest at -30.0 m/s (a) total vector displacement (Enter the magnitude in m and the direction in degrees south of west.) magnitude direction south of west (b) average speed (in m/s n/s (c) average velocity (Enter the magnitude in m/s and the direction in degrees south of west magntude direction mys south of west

Explanation / Answer

A)

S = S1 + S2 + S3

S = (-24 x 3 x 60)j - (25 x 3 x 60)i - (30 x 60 cos45)i + (30 x 60 sin45)j

= - 4320j - 4500i - 1272.79i + 1272.79j

= - 5772.79i - 3047.21j

Magnitude, S = sqrt (5772.79^2 + 3047.21^2) = 6527.68 m

Direction, theta = arctan (3047.21/5772.79) = 27.83 degree

B)

Average speed = total distance /total time taken

= [(24 x 3 x 60) + (25 x 3 x 60) + (30 x 60)]/(180 + 180 + 60)

= 25.29 m/s

C)

Magnitude = s/t = 6527.68/(180 + 180 + 60) = 15.54 m/s

Direction = 27.83 degree

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