Question

Janelle has 4 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the race and walking back the same distance to cool down. Of she runs at a speed of 7 mph and walks back at a speed of 3 mph, how long should she plan to spend walking back?

Answer 1

I am assuming one needs to find the distance of the race?

Answer 2

no

Answer 3

Janelle has 4 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the race and walking back the same distance to cool down. Of she runs at a speed of 7 mph and walks back at a speed of 3 mph, how long should she plan to spend walking back?

Answer 4

Okay, to solve this problem we need to figure out what portion of her time she spends running and what portion she spends walking. Janelle runs at 7mph and walks at 3mph, so he running speed is 7/3 her walking speed. Then any amount of time she spends running, she must spend 7/3 that much time walking back. Thus, we determine that if she spends x hours running, she spends 7x/3 hours walking, and x + 7x/3 = 4. Then we get 10x/3 = 4, or x = 6/5. Thus, Janelle spends 6/5 hours, or 1 hour and 12 minutes running and her remaining time, 2 hours and 48 minutes (which is 7/3 of 1 hour 12 minutes), walking

Answer 5

you are solving for the time she will spend walking back, first lets get some formulas down. Let d be the distance of the track, r1 and t1 be her speed and time running, r2 and t2 be he speed walking. Then we have:
\[d = r_{1}t_1{}, d=r_{2}t_{2}, t_1+t_2 = 4 \]
note: since rate is given in mph, t will be measured in hours.
Since the distance she will walk and run is the same, setting them equal to each other will give:
\[7t_1 = 3 t_2, t_1+t_2 = 4\]
Thus we have a system of equations to solve. t1 = (3/7)t2, so putting that in the second equation:
\[\frac{3}{7}t_2+t_2 = 4 \Rightarrow t_2 = \frac{4}{\frac{10}{7}} = 2.8 hours\]