Physics Help?

Question

A golfer rides in a golf cart at a speed of 3.10 m/s for 30.0 s. She then gets out of the cart and starts walking at an average speed of 1.50 m/s. For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 2.50 m/s?

Answer 1

I already answered this but here it is again. If you do give points please give them to whoever you decide for BOTH questions.

Left V be the average speed
v0 is the speed of the cart for a time t0
v1 is the walking speed for a time t1
T is the total time both in the cart and walking

V = = 2.5 m/s
v0 = 3.10 m/s and t0 = 30s
v1 = 1.5 m/s and t1 = Unknown
T = Unknown = 30 + t1

If Distance is the total distance she travels then the average speed is just this divided by the total time it takes her. So:
V = Distance/T

Distance is just the sum of the distances moved in the cart and in walking so:
Distance = v0*t0 + v1*t1
2.5 = [(3.1)(30) + 1.5t1] / (30 + t1)

Solve for t1:
2.5(30 + t1) = (3.10)(30) + 1.5t1
75 + 2.5t1 = 93 + 1.5t1
t = 93 - 75 = 18 seconds

She must walk for 18 seconds to give an average speed of 2.5 m/s.

Check:
d1 = cart distance = 3.1(30) = 93 m
d2 = walking distance = 1.5(18) = 27 m
d = d1 + d2 = 93 + 27 = 120 m
T = t0 + t1 = 30 + 18 = 48 s
V = d/T = 120/48 = 2.5 m/s