# Thread: Constant Velocity Problem Restated

1. ## Constant Velocity Problem Restated

This is a repost. I had screwed up the original posting, which now appears as a double-posted message if you just look at the subject line of the two messages just below this one, with only one of the posts being complete. Here is the entire post:

This problem came from my physics book, but the pre-calculus forum seems the most appropriate, as there is no physics forum, unless I missed it. I was able to solve the problem, but in a way that was not true to the description, and am wondering what the algebraic solution is. This is the problem description:

A person takes a trip, driving with a constant speed of 89.5 km/hr except for a 22.0-min rest stop. If the person's average speed is 77.8 km/hr, how much time is spent on the trip and how far does the person travel?
The way I solved this was by using two functions, one with a slope of 77.8 km/hr, with x and y intercept at the origin, and the other with a slope of 89.5 km/hr with an x-intercept of (.367 hr, 0 km), and a y-intercept of (0 hr, -32.8 Km), this y intercept computed by figuring how much displacement occurs in 22 minutes at 89.5 Km/hr. So my two equations were y = 77.8x and y = 89.5x - 32.8. Plotting them on a TI-89 and noting the intersection (2.8 hr, 218 Km), I arrived at the correct answer.

However, how to figure this out algebraicly is stumping me, and the book does not have a worked-out example like this problem. This is very frustrating, as I feel that I'm missing something very basic here.

2. what more algebra is there to it :P, once youve figured out what kinda displacement the break causes as you did, you have two equations both equal to y, as you stated, so? set them equal to each other and solve for x(the number of hours) and then finish by solving for the distance

good work btw

3. Originally Posted by spiritualfields
This is a repost. I had screwed up the original posting, which now appears as a double-posted message if you just look at the subject line of the two messages just below this one, with only one of the posts being complete. Here is the entire post:

This problem came from my physics book, but the pre-calculus forum seems the most appropriate, as there is no physics forum, unless I missed it. I was able to solve the problem, but in a way that was not true to the description, and am wondering what the algebraic solution is. This is the problem description:

The way I solved this was by using two functions, one with a slope of 77.8 km/hr, with x and y intercept at the origin, and the other with a slope of 89.5 km/hr with an x-intercept of (.367 hr, 0 km), and a y-intercept of (0 hr, -32.8 Km), this y intercept computed by figuring how much displacement occurs in 22 minutes at 89.5 Km/hr. So my two equations were y = 77.8x and y = 89.5x - 32.8. Plotting them on a TI-89 and noting the intersection (2.8 hr, 218 Km), I arrived at the correct answer.

However, how to figure this out algebraicly is stumping me, and the book does not have a worked-out example like this problem. This is very frustrating, as I feel that I'm missing something very basic here.
Your method is ok, as far as accuracy is concerned, but it's an overkill, here's the "basic" algebra you will need. someone else might have a different method, but this is how i thought of it

recall that $\mbox { Speed } = \frac { \mbox { Distance } }{ \mbox {Time} }$. we can describe each of the two speeds given this way to form a system of simultaneous equations to solve for the unknowns

Let $d$ be the total distance travelled
Let $x$ be the total time taken (this includes the time for the rest stop, which is 0.367 hours)

The average speed is the total distance traveled over the total time taken, so we have:

$77.8 \mbox { km/h } = \frac {d}{x}$

$\Rightarrow 77.8x - d = 0$ .................(1)

The constant speed that the car travels is the distance traveled over the time the car is actually traveling. this time is the total time minus the time for the rest stop. so we have:

$89.5 \mbox { km/h } = \frac {d}{x - 0.367}$

$\Rightarrow 89.5x - d = 32.847$ ......................(2)

So we obtain the system:

$77.8x - d = 0$ ..........................(1)
$89.5x - d = 32.847$ ...................(2)

Solving this system for $x$ and $d$ will give you the required time and distance respectively. The answers end up being the same as yours, but no graphs or TI-89 needed

EDIT: hmm, incidentally, those are the same two equations you came up with, except for the x and d as the variables of course. but your method involves graphing, mine is just pure algebra, which is what i think you are after

4. hmm, incidentally, those are the same two equations you came up with, except for the x and d as the variables of course.
Yes, the basic difference is I use y for the distance instead of d. Solving this via simultaneous equations is even more involved than what I was looking for. Perhaps there is no other way to solve this problem algebraicly, given the information provided, although matrices would work also. I was wondering if I was just too stupid to see an easier solution or what. I'm only reading the physics book, and not painstakingly working out each problem, and expected exercises that were at least in the ballpark of the worked-out examples. The worked out examples only required plugging values into the constant acceleration equations or finding an unknown in v = d/t. Real simple stuff. Maybe this is why the book I'm using (College Physics by Serway and Faughn) has such a poor rating on Amazon.com.

5. A person takes a trip, driving with a constant speed of 89.5 km/hr except for a 22.0-min rest stop. If the person's average speed is 77.8 km/hr, how much time is spent on the trip and how far does the person travel?
Let total time be $t$ hours, and distance be $d$ kilometres. Then the average speed for the journey is:

$
\frac{d}{t}=77.8
$

and the average speed while moving is:

$
\frac{d}{ t-\frac{22}{60} }=89.5
$

Divide the first by the second to get:

$
\frac{t-\frac{22}{60}}{t}=\frac{77.8}{89.5}
$

so:

$
t-\frac{22}{60}=\frac{77.8}{89.5}~t
$

Which you solve for $t$, then use one of the earlier equations to solve for $d$

RonL

6. Thanks, Captain Black. Pretty nifty. It looks like you used the principle of equals divided by equals are equal, and in the process the d variable is canceled out.

7. Originally Posted by spiritualfields
Thanks, Captain Black. Pretty nifty. It looks like you used the principle of equals divided by equals are equal, and in the process the d variable is canceled out.
Yes, sort of. It is actualy the other way around - get the equations then look
for the easiest way of eliminating one variable.

RonL

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# A person takes a trip driving with a constant speed of 89.5km/h except for a 22minute rest stop. if the persons average speed is 77.8km/h. calculate how much time is spent on the trip and how far does the person travel

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