Question

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. A particular small car, with a mass of 920 kg, can accelerate on a level road from rest to $21\mathrm{m}/\mathrm{s}\left(75\mathrm{km}/\mathrm{h}\right)$in 12.5 s. Using these data, calculate the maximum steepness of a hill.

Short answer

The maximum steepness of the hill is $9.9^{\mathrm{o}}$.

Step-by-step solution

Step 1. Statement of Newton’s second law of motion

Newton’s second law of motion states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.

Mathematically,

$a\propto \frac{F_{\mathrm{net}}}{m}$

Step 2. Identification of given data

The mass of the car is $m=75\mathrm{kg}$.

The initial velocity of the car is $u=0$.

The final velocity of the car is $v=21\mathrm{m}/\mathrm{s}$.

The time taken is $t=12.5\mathrm{s}$.

Step 3. Drawing an FBD for the car climbing the hill

A free body diagram of the car climbing up the hill is drawn below.

The forces acting on the moving car are:

• Normal force, N exerted upward on the car
• Weight, W of the car acting downward
• Force, F exerted to push the car against the road

Step 4. Determination of the acceleration of the car on a level road

Using the following kinematic equation of motion to determine the acceleration of the car on level ground, you get:

$\begin{array}{c}v=u+at\\ a=\frac{v-u}{t}\\ =\frac{21\mathrm{m}/\mathrm{s}-0}{12.5\mathrm{s}}\\ =1.68\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 5. Determination of the external force required to push the car

From Newton’s second law of motion, the force pushing the car forward to have acceleration is given by:

$\begin{array}{c}F=ma\\ =\left(920\mathrm{kg}\right)\left(1.68\mathrm{m}/{\mathrm{s}}^2\right)\\ \approx 1546\mathrm{N}\end{array}$

Step 6. Determination of the maximum angle or steepness for the hill

It is given that the car does not slow down while climbing the hill. Assume that the car has a constant speed with zero acceleration. The force, F, will be equal to the magnitude of the weight component directed down the hill.

From Newton’s second law of motion,

$\begin{array}{c}F=mg\sin \theta \\ \sin \theta =\frac{F}{mg}\\ \theta ={\sin }^{-1}\left(\frac{F}{mg}\right)\end{array}$

Substitute the numerical values in the above expression for the maximum angle of the hill.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1546\mathrm{N}}{\left(920\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\right)\\ =9.9^{\mathrm{o}}\end{array}$

Thus, the maximum angle for the hill is $9.9^{\mathrm{o}}$.