A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.

a. The speed of the race car
b. The distance traveled
c. the elapsed time

Question

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okpalawalter8 okpalawalter8 · Answer 1 · 2021-06-23T02:25:29+02:00

Answer:

a) [tex]V=14.904m/s[/tex]

b) [tex]d = 191.49 m[/tex]

c) [tex]t= 25.696 s[/tex]

Explanation:

From the question we are told that:

Radius [tex]r =378m[/tex]

Acceleration [tex]a=0.580[/tex]

a)

Generally the equation for speed of the car is mathematically given by

[tex]a=\frac{v^2}{r}[/tex]

[tex]V=\sqrt{a*r}[/tex]

[tex]V=\sqrt{0.58*383}[/tex]

[tex]V=14.904m/s[/tex]

b)

Generally the equation for distance traveled of the car is mathematically given by

[tex]V^2=u^2+2ad[/tex]

[tex]d=\frac{V^2}{2a}[/tex]

[tex]d=\frac{14.904^2}{2*0.58}[/tex]

[tex]d = 191.49 m[/tex]

c)

Generally the equation for time of the car is mathematically given by

[tex]V=u+at[/tex]

[tex]t=\frac{V}{a}[/tex]

[tex]t=\frac{14.904}{0.58}[/tex]

[tex]t= 25.696 s[/tex]