A car exhibits a constant acceleration of 0.200 m/s2 parallel to the roadway. Th

Question

A car exhibits a constant acceleration of 0.200 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 530 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 8.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant?

Conceptualize Conceptualize the situation using the figure and any experiences you have had in driving over rises on a roadway.

Categorize Because the accelerating car is moving along a curved path, we categorize this problem as one involving a particle experiencing both tangential and radial acceleration. We recognize that it is a relatively simple substitution problem.

Analyze The radial acceleration is given by the following formula, with v = 8.00 m/s and r = 530 m. The radial acceleration vector is directed straight downward, and the tangential acceleration vector has magnitude 0.200 m/s2 and is horizontal.
Evaluate the radial acceleration:
ar = -v2/r= m/s2

Find the magnitude of :
a = ?ar2 + at2 = m/s2

Find the angle ? (see figure) between and the horizontal:
? = tan-1(ar / at) =

A car is speeding up uniformly at a rate of 0.700 m/s2 while traveling counterclockwise around a circular track of radius 475 m. At the instant the car is moving due north, its speed is 10.00 m/s. What is the magnitude and direction of the total acceleration of the car at this instant? Use a positive angle for west of north and a negative angle for east of north.

magnitude m/s2

direction

Explanation / Answer

radial acc = v^2 / r = 8^2/530 = 0.121
tangential acc = 0.2
total acc = (0.121^2 + 0.2^2) =0.234 m/s^2

direction = tan^-1 (0.121/0.2) = 31.174

radial acc= v^2/r = 10^2 / 475 = 0.21 west

tangential acc = 0.7 north

total acc= (0.7^2 +0.21^2) = 0.731

direction = tan^-1(0.21/0.7) =16.7 west of north

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