# car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway

Question

car leaves a stop sign and exhibits a constant acceleration of 0.300 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 an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant

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1 year 2021-09-04T13:32:11+00:00 1 Answers 225 views 0

0.308 m/s2 at an angle of 13.5° below the horizontal

Explanation:

The parallel acceleration to the roadway is the tangential acceleration on the rise.

The normal acceleration is the centripetal acceleration due to the arc. This is given by

$$a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072$$

The tangential acceleration, from the question, is

$$a_T = 0.300$$

The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by

$$a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2$$

$$a = 0.308$$

The angle the resultant makes with the horizontal is given by

$$\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400$$

$$\theta=13.5$$

Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc