Question

!50 POINTS! GIVE REAL ANSWER OR I REPORT PLZ NO TROLLS! An executive drove from his home at an average speed of 35 mph to an airport where a helicopter was waiting. The executive then boarded the helicopter and flew to the corporate offices at an average speed of 78 mph. The entire distance from his home to the office was 169.5 miles. If the executive spent the same amount of time in the car as he did the helicopter, how long did it take the executive to get to work?

Answer 1

Answer:

it took him 3 hours to get to work

Step-by-step explanation:

For the first part of the trip, the executive covered a distance "d" in time 't" at 35 mph, therefore we can write that the time it took to get to the helicopter was:

$t-\frac{d}{35}$

which if the distance is in miles, will render the time in hours.

We can write something similar for the second part of the trip", using the same "t" since it took the same time as for the first one, and using for distance 169.5 - d (since it is the distance to complete the total traveled distance :

$t=\frac{169.5-d}{78}$

Now we equal those two equations because the time "t" is the same, and solve for "d":

$\frac{d}{35} =\frac{169.5-d}{78} \\78\,d=35\,(169.5-d)\\78\,d=5932.5-35\,d\\(78+35)\,d=5932.5\\d=\frac{5932.5}{113} \\d=52.5$

Now, knowing the distance covered in the first trip, we can find the time of that first trip:

$t=\frac{52.5}{35} =1.5\,\,hours$

Therefore, the time it took for thye total trip is: 1.5 h + 1.5 h = 3 hours