A runner in a circular track was observed running at an angular velocity of 0.23

Question

A runner in a circular track was observed running at an angular velocity of 0.230 rad/s. 2.4 seconds later, she was observed running at 0.050 rad/s. The radius of the circular track is 61m Show work!!!

a) What is in radians/second2 her angular acceleration during the 2.5 seconds?

b) What is her speed in meters per second at the end of the 2.5 seconds?

c)  What is  in meters/second2 her tangential acceleration during the 2.5 seconds?

d)  What is  in meters/second2 her centripetal acceleration at the end of  the 2.5 seconds?

Explanation / Answer

Initial angular velocity w= 0.230 rad/s

Time t = 2.4 seconds

Final angular speed w ' = 0.050 rad/s

The radius of the circular track r = 61m

a)Angular acceleration during the 2.5 seconds a = ( w ' - w) / t

                                                                      = (0.05-0.23) /2.4

                                                                       = -0.075 rad/s 2

b) Her speed in meters per second at the end of the 2.5 seconds is v =?

Angular speed after 2.5 s is w " = w + at '

                                              = 0.23 +(-0.075 x2.5)

                                              = 0.0425 rad/s

So, v = rw"

        = 2.59 m/s

c) Her tangential acceleration during the 2.5 seconds is a ' = ( v - u ) / t '

Where u = initial speed = rw

              = 61 x 0.23

              = 14.03 m/s

a ' = -4.576 m/s 2

d) Her centripetal acceleration at the end of  the 2.5 seconds is = v 2 / r

                    = 0.11 m/s 2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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