Kinematics - velocity of golfer

  • Thread starter RKNY
  • Start date
  • #1
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Problem 1

Homework Statement


A golfer rides in a golf cart at a speed of 3.10 m/s for 29.0 s. She then gets out of the cart and starts walking at an average speed of 1.20 m/s. For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 2.10 m/s?


Homework Equations


time = distance x avg. speed


The Attempt at a Solution


Taking the distance traveled from the riding -
3.10 m/s x 29.0 s = 90 m
Tried numerous ways such as taking another 90m and using the 1.20 as a given



Problem 2

Homework Statement


A sprinter explodes out of the starting block with an acceleration of +3.5 m/s2, which she sustains for 1.2 s. Then, her acceleration drops to zero for the rest of the race.

(a) What is her velocity at t = 1.2 s.
(b) What is her velocity at the end of the race?

Homework Equations


Avg. Veloctiy = displacement/elapsed time
Avg. Accel. = Change in veloc./elapsed time

The Attempt at a Solution


(a) 3.5 m/s^2 x 1.2 s = 4.2 m/s
(b) I believe that you would have to use -3.5 because of the decrease in acceleration but I can't seem to know how to get anything or figure anything out after that.
 

Answers and Replies

  • #2
960
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Think about the two components of her travel distance
1) in cart = 3.1*29s
2) while walking 1.2*t (t=unknown time that we wish to solve)

Then the total distance which can be expressed as the total time(29+t)*ave speed. Equate the two, and solve for t.

Problem 2: You're maybe looking for something thats not there. Think about riding a bike, you crank furiously for 1.2 sec then abruptly stop pedaling. Now imagine doing so where there is no air resistance to slow you down and the rolling resistance from the tires negligible. what happens to the speed of the bike?
 

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