Kinematics - velocity of golfer

[RKNY]

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.

 

[denverdoc]

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 that's 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?

 

[m2003]

For Problem 1, we can calculate the total distance traveled by the golfer by adding the distance traveled while riding (90 m) and the distance traveled while walking (1.2 m/s x t). Since we know that the average speed for the entire trip is 2.10 m/s, we can set up the equation:

2.10 m/s = (90 m + 1.2 m/s x t) / (29.0 s + t)

Solving for t, we get t = 10.0 seconds. Therefore, the golfer must walk for 10.0 seconds to maintain an average speed of 2.10 m/s for the entire trip.

For Problem 2, we can use the equation for average acceleration to calculate the final velocity at the end of the race:

Avg. Accel. = Change in velocity / elapsed time

Since the acceleration drops to zero after 1.2 seconds, we can set up the equation:

0 = (v - 4.2 m/s) / (t - 1.2 s)

Solving for v, we get v = 4.2 m/s. Therefore, the sprinter's velocity at the end of the race is 4.2 m/s.

For part (a), her velocity at t = 1.2 s, we can use the equation for average velocity:

Avg. Velocity = displacement / elapsed time

Since her acceleration is constant at 3.5 m/s^2, we can use the equation for displacement:

Displacement = initial velocity x elapsed time + 1/2 x acceleration x (elapsed time)^2

Plugging in the values, we get:

v = 0 m/s + 1/2 x 3.5 m/s^2 x (1.2 s)^2 = 2.52 m/s

Therefore, her velocity at t = 1.2 s is 2.52 m/s.