Answer:
[tex]\beta =32.313[/tex]
Explanation:
The first step is take all the information in the same units so the speed is going to pass in ft/s^2
[tex]v=98 \frac{mi}{h}*\frac{1h}{60minute}*\frac{1minute}{60s}=0.0272222\frac{mi}{s}[/tex]
[tex]v=0.027222\frac{mi}{s}*\frac{5280ft}{1mi}=143.73 \frac{ft}{s}[/tex]
Now the velocity for a oval racetrack is
[tex]v=\sqrt{r*g*tan(\beta)}[/tex]
β= the angle necessary for the race cars at 98 mi/h so
[tex]v^{2}=r*g*tan(\beta)[/tex]
[tex]tan(\beta)=\frac{V^2}{r*g}[/tex]
[tex]tan(\beta)=\frac{(142.26 \frac{ft}{s})^{2}}{1000ft*32\frac{ft}{s^2}}[/tex]
[tex]\beta =tan^-1*(0.6324)[/tex]
[tex]\beta =32.313[/tex]
The banking angle theta necessary for the race cars to navigate the turns at 98 mi/h without the aid of friction is 32.86 degrees
The formula for finding the banking angle theta necessary for the race cars to navigate is expressed according to the equation:
[tex]v = \sqrt{rg tan \alpha}[/tex] where:
v is the velocity = 98mi/hr = 143.73ft/s
g is the accceleration due to gravity = 32ft/s²
[tex]\alpha[/tex] is the banking angle
r is the radius = 1000ft
Make the banking angle the subject of the formula;'
[tex]v^2 =rg tan\alpha\\tan\alpha = \frac{v^2}{rg}\\tan\alpha = \frac{143.73^2}{1000\times 32}\\tan\alpha = \frac{20,658.3129}{32000}\\tan\alpha =0.6455\\\alpha = arctan0.6455)\\\alpha = 32.86[/tex]
Hence the banking angle theta necessary for the race cars to navigate the turns at 98 mi/h without the aid of friction is 32.86 degrees
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