physics help please?

Question

A golfer rides in a golf cart at a speed of 3.40 m/s for 35.0 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 2.60 m/s? (answer is in units of s)

Answer 1

For her to have a final average speed of 2.60 m/s, she must travel a total distance X in a total time T such that X/T = 2.60 m/s. If she rides at 3.40 m/s for 35.0 s, she has traveled 119 m. If she continues at 1.50 m/s for some additional time t (your unknown), she would end with a total X = 119 + 1.50*t and a total T = 35.0 + t. So we need to solve (119 + 1.50*t) / (35.0 + t) = 2.60 ==> (119 + 1.5t) / (35 + t) = 2.6 ==> 119 + 1.5t = 2.6(35 + t) ==> 119 + 1.5t = 91 + 2.6t ==> 28 = 1.1t ==> t = 25.45 s.

Answer 2

Let her walk for t seconds. Then,
The total distance S travelled: S = 3.40*35 +1.50 t ( in meters)
Total time T taken : T=(35 +t) s
Average speed v : v = ( 3.40*35 + 1.50 t) / (35 + t).
This is given to be 2.60 m/s. So:
( 3.40*35 + 1.50 t) / (35 + t). =2.60 Or
3.40*35 + 1.50 t =2.60*35 +2.60 t OR
(3.40 -2.60)*35 = (2.60 - 1.50) t OR
0.80 *35= 1.1 t OR t = 0.80* 35/1.1 = 25.5 s Answer.

Answer 3

avg V = displacement / total time


total displacement 119 + x
total time = 35 sec + x/1.5

put in equation and solve

Answer 4

3.4 x 35+1.5x=2.6x+35(2.6)
119 + 1.5x = 2.6x + 91
-1.1x = -28
x = 25.4545454545s

Answer 5

seriously, this is the second homework question you've asked... at the least, give me the best answer on the last one i did for you... :P

Answer 6

write an equation.

3.4x=35
1.5x=2.6y

Answer 7

Do your own homework....