A car traveling at a speed of 20m/s due north along the way makes a right turn on to a side road that heads to east. It takes 50 sec for the car to complete the turn. At the end of the 50s, the car has the speed of 15m/s along the side road. Determine the magnitude of average acceleration over the 50s interval.

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Solution

The car initially travls at 20 j (m/s).

At the end of 50 seconds the velocity becomes 15 i.

Therefore acceleration along x = 15-0/time = 15/50 = 0.3 i

Acceleration along y = 0-20/time = -20/50= -0.4j

Thus th total acceleration is 0.3i +0.4j.

thus the magnitude is 0.5

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The car initially travls at 20 j (m/s). At the end of 50 seconds the velocity becomes 15 i. Therefore acceleration along x = 15-0/time = 15/50 = 0.3 i Acceleration along y = 0-20/time = -20/50= -0.4j Thus th total acceleration is 0.3i +0.4j. thus the magnitude is 0.5

The car initially travls at 20 j (m/s).

At the end of 50 seconds the velocity becomes 15 i.

Therefore acceleration along x = 15-0/time = 15/50 = 0.3 i

Acceleration along y = 0-20/time = -20/50= -0.4j

Thus th total acceleration is 0.3i +0.4j.

thus the magnitude is 0.5