Question

An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions, with origin at the center of the circle. Suppose the vehicle starts from rest at x = 50 m heading north, and its speed depends on the distance s it travels according to v = 0.5s − 0.0025s 2 , where s is measured in meters and v is in meters per second. It is known that the tires will begin to skid when the total acceleration of the vehicle is 0.6g. Where will the automobile be and how fast will it be going when it begins to skid? Describe the position in terms of the angle of the radial line relative to the x axis.

Answer 1

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ($a$) can be defined by this ordinary differential equation in terms of distance:

$a = v\cdot \frac{dv}{ds}$ (1)

Where:

$v$ - Speed of the automobile, measured in meters per second.

$s$ - Distance travelled by the automobile, measured in meters.

If we know that $v = 0.5\cdot s - 0.0025\cdot s^{2}$, then the equation of acceleration is:

$a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)$

$a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)$

$a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)$

But distance covered by the vehicle is defined by the following formula:

$s = \theta \cdot r$ (2)

Where:

$\theta$ - Arc angle, measured in radians.

$r$ - Radius, measured in radians.

Then, we expand (1) by means of this result:

$a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)$

$a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)$

$a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r$

And finally we get the following third order polynomial:

$1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0$ (3)

If we know that $r = 50\,m$, $a = 0.6\cdot g$ and $g = 9.807\,\frac{m}{s^{2}}$, then the polynomial becomes into this:

$1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0$ (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

$\theta_{1} \approx 4.365\,rad$, $\theta_{2} \approx 0.818+i\,0.441\,rad$, $\theta_{3}\approx 0.818 -i\,0.441\,rad$, $\theta_{4} \approx 1.563\,rad$

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ($r = 50\,m$, $\theta \approx 1.563\,rad$)

$s = (1.563\,rad)\cdot (50\,m)$

$s = 78.15\,m$

Lastly, the speed of the automobile at this location is: ($s = 78.15\,m$)

$v = 0.5\cdot s - 0.0025\cdot s^{2}$ (4)

$v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}$

$v = 23.806\,\frac{m}{s}$

The automobile is running at speed of 23.806 meters per second.