For a perfectly rolling tired of 0.8 m radius the center of the tired makes an a

Question

For a perfectly rolling tired of 0.8 m radius the center of the tired makes an acceleration of 1.6 m/se .find the distance of the center of the tire moving from the rest in 3 minutes and number of revolution it turns For a perfectly rolling tired of 0.8 m radius the center of the tired makes an acceleration of 1.6 m/se .find the distance of the center of the tire moving from the rest in 3 minutes and number of revolution it turns For a perfectly rolling tired of 0.8 m radius the center of the tired makes an acceleration of 1.6 m/se .find the distance of the center of the tire moving from the rest in 3 minutes and number of revolution it turns

Explanation / Answer

a = 1.6 m/s^2

S = ut + 1/2 at^2

S = 0 + (.5 x 1.6 x (3 x 60)^2)

S = 25920 m (Distance)

1 revolution = 2 pi r = 2 x pi x .8 = 5.026 m

By simple unitary method

5.026 m = 1 revolution

1 m = 1 / 5.026 revolution

25920 m = (1 / 5.026) x 25920 revolution

Revolution = 5157.18 revolutions

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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