Lesson: Introduction to Motion Questions

Comment on Introduction to Motion Questions

https://gre.myprepclub.com/forum/the-time-required-to-travel-d-miles-at-s-miles-per-hour-8678.html
In this case we're so sure speed is positive because speed can never be negative?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/the-time-required-to-travel-d-miles-at-...

That's correct. The speed is always greater than or equal to zero.

Cheers,
Brent

ds not equal to zero means both numbers are either positive or negative?
greenlight-admin's picture

Not quite.
If ds ≠ 0, then neither d nor s are equal to zero.

So, for example, if xy ≠ 0, then it could be the case that x > 0 and y < 0.
Or, it could be the case that x > 0 and y > 0
Or, it could be the case that x < 0 and y > 0
Or, it could be the case that x < 0 and y < 0

Cheers,
Brent

It is basically the same thing, but I prefer writing out a series of ratios and then cancelling the units. For me, it is easier to keep track of things that way. For example: https://imgur.com/a/DjwTRXP
greenlight-admin's picture

That strategy definitely works!

Harper began driving from R to G at the same time as Quinn began driving from G to R on the same road. Harper and Quinn met on the road 3 hours later. If Quinn traveled at a constant of 1/4 greater than Harper's constant rate and the distance between R and G is 270 miles what was Harper's rate.

So I approached this with the equation D of R-G + D G-R = 270 , I got the right answers that way, but wanted to know if its possible to solve with equation, I set up as Harper's Time + Quinn's Time = 3
270/Hr + 270/(1.25Hr) = 3 and solved for Hr which was not the same value as I did before. Is my equation wrong
greenlight-admin's picture

There are a couple of errors with that approach.
First, since the two people meet three hours after they started, EACH person's travel time is 3 hours.
So we actually get two equations:
Harper's time = 3
Quinn's time = 3

There's also an issue with your equation 270/Hr + 270/(1.25Hr) = 3
This equation suggests that each person travels 270 miles, but this isn't the case.
The COMBINED distance each person travels = 270

So here's how we can set up the solution....

Let R = Harper's speed
So, 1.25R = Quinn's speed

Let D = The distance Harper traveled
So, 270 - D = the distance Quinn traveled

We can all take the equation: Harper's time = 3
And substitute to get: D/R = 3
Multiply both sides by R to get: D = 3R

Now take the other equation: Quinn's time = 3
And substitute to get: (270 - D)/1.25R = 3
Multiply both sides by 1.25R to get: 270 - D = 3.75R

We now have the following system of equations:
D = 3R
270 - D = 3.75R

When we solve the system we get R = 40 (mph)

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