Chapter 8.4, Problem 5MRPS

Solution

(a)

To calculate : The radius of the skid mark shown in the figure.

The radius of the skid mark is $58.6$ meters.

Given information :

A car skids while turning to avoid an accident. The car's speed $v$ (in meters per second) can be estimated using $v=\sqrt{9.8\mu r}$ where $r$ is the radius(in meters) of the circular skid mark and $\mu$ is a constant that depends on the road surface and weather conditions.

  

Calculation :

The skid mark can be compared to a translated circle. The equation of a translated circle with center at $\left(h,k\right)$ and radius $r$ is given by,

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Now, for the point $\left(0,0\right)$ , the equation will be,

  $\begin{array}{l}{\left(0-h\right)}^2+{\left(0-k\right)}^2=r^2\\ h^2+k^2=r^2\mathrm{.......}\left(1\right)\end{array}$

For the point $\left(-10,1\right)$ , the equation will be,

  $\begin{array}{c}{\left(-10-h\right)}^2+{\left(1-k\right)}^2=r^2\\ 100+20h+h^2+1+k^2-2k=r^2\\ h^2+k^2+100+20h+1-2k=r^2\\ r^2+101+20h-2k=r^2\left[\text{From }\left(1\right)\right]\\ 20h-2k+101=0\mathrm{.........}\left(2\right)\end{array}$

For the point $\left(16,2\right)$ , the equation will be,

  $\begin{array}{c}{\left(16-h\right)}^2+{\left(2-k\right)}^2=r^2\\ 256-32h+h^2+4+k^2-4k=r^2\\ h^2+k^2+256-32h+4-4k=r^2\\ r^2+260-32h-4k=r^2\left[\text{From }\left(1\right)\right]\\ -32h-4k+260=0\\ 8h+k-65=0\left[\text{Take }4\text{ as common}\right]\\ k=65-8h\end{array}$

Substitute the value of $k$ in (2).

  $\begin{array}{c}20h-2\left(65-8h\right)+101=0\\ 20h-130+16h+101=0\\ 36h=29\\ h\approx 0.8\end{array}$

Substitute the value of $h$ in $k=65-8h$ .

  $\begin{array}{c}k=65-8\left(0.8\right)\\ =58.6\end{array}$

Substitute the value of $h$ and $k$ in (1)

  $\begin{array}{c}{r}^2={\left(0.8\right)}^2+{\left(58.6\right)}^2\\ =3434.6\\ r=58.6\end{array}$

Therefore, the radius of the skid mark is $58.6$ meters.

(b)

To calculate : The speed of the car for $\mu =0.7$ .

The required values are $a=36.5,\text{ and }c=7.5$ .

Given information :

A car skids while turning to avoid an accident. The car's speed $v$ (in meters per second) can be estimated using $v=\sqrt{9.8\mu r}$ where $r$ is the radius(in meters) of the circular skid mark and $\mu$ is a constant that depends on the road surface and weather conditions.

  

Calculation :

The radius of the skid mark is $58.6$ meters.

The speed of the car for $\mu =0.7$ is,

  $\begin{array}{c}v=\sqrt{9.8\times 0.7\times 58.6}\\ =\sqrt{401.996}\\ \approx 20\end{array}$

Therefore, the speed of the car is $20$ meters per second.

(c)

To explain : The effect of speed of car while skidding on the radius of the circles.

When the speed increases the radius also increases.

Given information :

A car skids while turning to avoid an accident. The car's speed $v$ (in meters per second) can be estimated using $v=\sqrt{9.8\mu r}$ where $r$ is the radius(in meters) of the circular skid mark and $\mu$ is a constant that depends on the road surface and weather conditions.

  

Explanation :

The speed of the car while skidding is given by the formula $v=\sqrt{9.8\mu r}$ .

This shows that the speed of the car is directly proportional to the radius.

Therefore, when the speed increases the radius also increases.