Number 22: A race car starts from rest on a circulartrack of radius 400m. The ca

Question

Number 22: A race car starts from rest on a circulartrack of radius 400m. The car's speed increases at theconstant rate of .500 m/s2. At that point where the magnitudeof the centripetal and tangential accelerations are equal,determine (a) the speed of the race car, (b) the distance traveled,and (c) the elapsed time. This is the full problem if anyone can help I would appreciateit Number 22: A race car starts from rest on a circulartrack of radius 400m. The car's speed increases at theconstant rate of .500 m/s2. At that point where the magnitudeof the centripetal and tangential accelerations are equal,determine (a) the speed of the race car, (b) the distance traveled,and (c) the elapsed time. This is the full problem if anyone can help I would appreciateit

Explanation / Answer

(a)       Given tangentialacceleration ar = 0.500 m/s2       We have ar = V2 / r       Velocity is V = ar * r                                 = 0.50 * 400                                 = 14.142135   m /s (b)        Distance travelled is:                    S = V` t   = (14.142135 / 2 )* 28.2             since V` = V /2                                      = 199.4041 m (c)        t   = V /ar = 14.142135   / 0.500                          = 28.2   s Hope this helps u!

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.