Question 4.7: Over the Rise A car leaves a stop sign and exhibits a consta...
Over the Rise
A 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. At the moment 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?
Conceptualize Conceptualize the situation using Figure 4.18a 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.
The tangential acceleration vector has magnitude 0.300 m/s² and is horizontal. The radial acceleration is given by Equation 4.28, with v = 6.00 m/s and r = 500 m. The radial acceleration vector is directed straight downward.
a_r=-a_c=-\frac{v^2}{r} (4.28)
Evaluate the radial acceleration:
a_r=-\frac{v^2}{r}=-\frac{(6.00 m/s)^2}{500 m}=-0.072 0 m/s^2Find the magnitude of \overrightarrow{a}:
\begin{aligned} a=\sqrt{a_r{}^2+a_t{}^2}&=\sqrt{\left(-0.072 0 m/s^2\right)^2+\left(0.300 m/s^2\right)^2} \\& =0.309 m/s^2\end{aligned}Find the angle Φ (see Fig. 4.18b) between \overrightarrow{a} and the horizontal:
\phi=\tan ^{-1} \frac{a_r}{a_t}=\tan ^{-1}\left(\frac{-0.072 0 m/s^2}{0.300 m/s^2}\right)=-13.5^{\circ}