Question

please help me with the finding out the variables and the equation(s) to be used, more than the answer.. thanks so much

Edwin averaged 50 kph traveling from his home to school. Returning by a different route
that covered the same distance, he averaged a speed of 55 kph. If his total traveling time was
32 hours, then how far is his home from school?

Answer 1

Ok, you're gonna have several equations here, so let's get them out on the table. The equations have one thing in common: the distance covered.
Let d reference distance, t reference time, and v reference velocity.
For the first route:\[d_{1} = v_{1}t_{1}\]And for the second route:\[d_{2} = v_{2}t_{2}\]
The problem says the total time was 32 hours. For now, we'll let the total time be T:\[t_{1}+t_{2} = T\]
The problem says both routes covered the same distance so we'll just call it d:\[d_{1} = d_{2} = d\]Rewrite the equations with d:\[d = v_{1}t_{1}\]\[d= v_{2}t_{2}\] Now you can substitute d in one of the equations with the other equation, or in other words, they both equal the same thing (d) so the equations must be equal:\[t_{1}v_{1} = t_{2}v_{2}\]
Using the relationship between t1 and t2, you can solve for one of them:\[t_{1} = T - t_{2}\]Now substitute t1 in the equation:\[(T-t_{2})v_{1} = t_{2}v_{2}\]Now you can solve for t2:\[t_{2} = \frac{v_{1}T}{v_{1}+v_{2}} = \frac{(50k/h)(32h)}{(50k/h)+(55k/h)} = \frac{320}{21}h \approx 15.24h\]Now that you know t2, you can solve for t1:\[t_{1} = T-t_{2} = 32h - \frac{320}{21}h = \frac{352}{21}h \approx 16.76h\]Now just plug in either of these time values into their respective equations to solve for d:\[d = t_{1}v_{1} = \left(\frac{352}{21}h\right) \left( 50k/h\right) \approx 838.095k\]\[d = t_{2}v_{2} = \left(\frac{320}{21}h\right) \left( 55k/h\right) = 838.095k\]

k is for kilometers I'm assuming

Answer 2

Wow, that was... comprehensive. Yes, k's for kilometers. Thank you so much! :)