Question 316580: A tourist can bicycle 52 miles in the same time as he can walk 20 miles. If he can ride 8 mph faster than he can walk, how much time should he allow to walk a 29 mile trail? (Hint: How fast can he walk?)
Answer by texttutoring(324)   (Show Source):
You can put this solution on YOUR website! We need to use the formula D=VT, where D=distance, V=velocity (or speed) and T=time
I will use subscript b for biking, and w for walking:
Db = 52 miles
Dw = 20 miles
Vb = Vw +8
The times are the same, so isolate for T in the D=VT equation:
Tw = Dw/Vw
Tb = Db/Vb
Now set Tb and Tw equal to each other:
Db/(Vw+8) = 20/Vw
Cross multiply and solve for Vw:
52*Vw = 20*Vw +160
32*Vw = 160
Vw = 5
So the man's walking speed is 5 mph.
To find the time to walk a 29 mile trail, use T=D/V
T = 29 / 5
T=5.8
He should allow 5.8 hours to walk a 29 mile trail.
|
|
|
