Chapter 1, Problem 4C

A spoils car is advertised to have a maximum cornering acceleration of 0.85 g.
(a) What is the maximum speed that the car can go around a curve with a 100-m radius?
(b) What is its maximum speed for a 50-m radius curve?
(c) If wet pavement reduces its maximum cornering acceleration to 0.6 g, what do the answers to (a) and (b) become?

To determine

(a)

Maximum speed of car around curve of radius 100m.

Answer to Problem 4C

Maximum speed of car around the curve of radius $100mis28.61m/s$.

Explanation of Solution

Given:

$\begin{array}{l}Accelerationinmultiplesof\text{ (}a)=0.85g\hfill \\ Radiusofthecurver=100m\hfill \end{array}$.

Formula used:

Acceleration is given by the formula,

$a=\frac{v^2}{r}$.

Calculation:

$\begin{array}{l}a=0.85g\\ where\text{ g = 9}{\text{.81 m/s}}^2\\ a=0.85\times 9.81\\ a=8.33m/s^2\end{array}$

We have value for a, now using formula,

$\begin{array}{l}a=\frac{v^2}{r}\\ v=\sqrt{ar}\\ v=\sqrt{8.33\times 100}\\ v=28.86m/s\end{array}$.

Conclusion:

The maximum speed of car on the curve of $100mis28.61m/s$.

To determine

(b)

The speed of a car around a curve of $50m$.

Answer to Problem 4C

The speed of car around a curve of $50mis20.40m/s$.

Explanation of Solution

Given:

$\begin{array}{l}Theradiusofthecurver=50m\hfill \\ Accelerationofcar=0.85g\hfill \end{array}$

Formula used:

Calculation:

Using the same formula above

$\begin{array}{l}a=8.33m/s^2\\ v=\sqrt{ar}\\ v=\sqrt{8.33\times 50}\\ v=20.40m/s\end{array}$.

Conclusion:

The speed of the car on a curve of $50mis20.40m/s$.

To determine

(c)

The speed of a car around a curve of $100m$ and around a curve of

$50m$ if acceleration is changed to 0.6g.

Answer to Problem 4C

The speed of car around a curve of $100mis24.24m/s$

The speed of car around the curve of $50mis17.14m/s$.

Explanation of Solution

Given:

$\begin{array}{l}Theradiusofthecurver=100m\hfill \\ Theradiusofthecurver=50m\hfill \\ Accelerationofcara=0.6g\hfill \end{array}$.

Calculation:

Similarly as above,

$\begin{array}{l}a=0.6g\\ a=0.6\times 9.81=5.88m/s^2\\ v=\sqrt{ar}\end{array}$

When r = 100m.

$\begin{array}{l}v=\sqrt{5.88\times 100}\\ v=24.24m/s\end{array}$

Now, when r = 50m.

$\begin{array}{l}v=\sqrt{5.88\times 50}\\ v=17.14m/s\end{array}$.

Conclusion:

$\begin{array}{l}Thespeedofthecaronacurveof100mis24.24\text{ m/s}\hfill \\ Thespeedofthecaronacurveof50mis17.14m/s\hfill \end{array}$.