Answer:
Acceleration, [tex]a=44.63\ m/s^2[/tex] Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Explanation:
It is given that,
Initial speed of the car, u = 0
Final speed of the car, v = 335 km/h = 93.05 m/s
Radius of the semicircular arc, r = 194 m
Distance covered by the car in the semicircular arc is, [tex]s=\dfrac{2\pi r}{2}=\pi r[/tex]
The tangential acceleration of the car is given by the following formula as :
[tex]a=\dfrac{v^2}{r}[/tex]
[tex]a=\dfrac{(93.05\ m/s)^2}{194\ m}[/tex]
[tex]a=44.63\ m/s^2[/tex]
So, the tangential acceleration of the car when it is halfway through the turn is [tex]44.63\ m/s^2[/tex]. Hence, this is the required solution.
The tangential acceleration of the car when it is halfway through the turn will be   [tex]a=44.63\dfrac{m}{s^2}[/tex]
Data we have
The initial speed of the car, u = 0
The final speed of the car, v = 335 km/h = 93.05 m/s
The radius of the semicircular arc, r = 194 m
Distance covered by the car in the semicircular arc is,
[tex]S=\dfrac{2\pi r}{2} =\pi r[/tex]
The formula for calculating tangential; acceleration will be
[tex]a=\dfrac{v^2}{r} = \dfrac{93.05^2}{194}[/tex]
[tex]a=44.63\ \frac{m}{s^2}[/tex]
Thus the tangential acceleration of the car when it is halfway through the turn will be   [tex]a=44.63\dfrac{m}{s^2}[/tex]
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