A sprinter runs at 8.3 m/s around a circular track with a centripetal accelerati

Question

A sprinter runs at 8.3 m/s around a circular track with a centripetal acceleration of magnitude 3.6 m/s2. This is a direct plug-in to the equations for centripetal acceleration and period. A train moving at a constant speed of 55.0 km/h moves east for 50.0 min, then in a direction 40.0 degree east of due north for 15.0 min, and then west for 40.0 min. What is the average velocity of the train during this run? (Let the +x-axis point towards the east and the +y-axis point towards the north.) To calculate an average velocity, divide the displacement by the time taken. That is, find the magnitude of the displacement vector pointing from start to finish and then divide by the time. (Don't divide the total distance by the time taken.) The displacement is the vector that points from start to finish. It is determined as in Chapter 3: Find the net displacement parallel to the east-west axis (an x axis) and the net displacement parallel to the north-south axis (a y axis). These two quantities are the x and y components of the displacement vector. Use the Pythagorean Theorem to find the magnitude of the vector and an inverse tangent to find an angle. Report the orientation relative to the positive direction of the x axis.

Explanation / Answer

a)

Centripetal acceleration

a=V2/r

=>r=V2/a =8.32/3.6

r=19.14 m

b)

Period of motion

T=2pir/V =2pi*19.14/8.3

T=14.5 seconds

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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