Speed/Velocity Problem

Homework Statement

A car travels at a speed of 30 km/h for a distance of 12 km. It then reaches a freeway where it travels at 85 km/h for 40 km. Find the average speed of the car for the entire trip.

Homework Equations

t=change in x over change in v.

The Attempt at a Solution

I know the equations for speed and velocity, but I can't figure out how to apply them to this problem. I made several attempts, which I erased. I don't want anyone to solve the problem for me, but I would really appreciate a list of steps and/or the directions for applying the formula.

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What do you need to know to find the average speed?

The only equation I was given was t=change in x/change in y. I apologize for the format of the equation, I have no clue how to make it look proper on here.

I assume your equation is $$t = \frac {\Delta x} {\Delta v}$$ Are you sure this is really the equation you were given? I would expect an equation like $$\Delta t = \frac {\Delta x} {v_{\text{average}}}$$

Yes, the top one is the one I referred to. It is the one I was given, but I was also told that the t could be exchanged with the v if needed. I turned back to my notes, and I also see the second equation recorded, but with only t images ebb of delta t. I had assumed the top and bottom equations were the same. I'm still in the first few weeks of my course, and am trying to smooth out a few details.

*with only t instead of delta t. Sorry. Auto-correct is out to get me.

Alright then.

You need the second equation to obtain the average speed, in which case it is best recast as $$v_{\text{average}} = \frac {\Delta x} {\Delta t}$$ (I am sticking with deltas for time and distance so that we know we are dealing with some intervals)

Now, to apply the equation and get your answer, you need to know both $\Delta x$ and $\Delta t$. Any idea about them?

Do the changes in x and t mean I should add the 30 km/h and the 40 km/h, and add the 12 km and 40 km?

Since you are looking for the average speed during the entire trip (as the problem states), $\Delta x$ means the change in distance during the entire trip. And $\Delta t$ simply means the duration of the entire trip. What are they?

Would delta x=52 km, and delta t=115 km/h ? Or vice-versa?

$\Delta x$ is 52 km indeed. However, $\Delta t$ is time, and 115 km/h, no matter how you got it, is not time, it is speed.

Great, thank you! So how do I figure out the time if I'm only given speed and distance?

Kot
Time can be found by dividing distance by speed.

Great, thank you! So how do I figure out the time if I'm only given speed and distance?
Your trip has two legs. You can compute the duration of each leg by applying the first formula.

Your trip has two legs. You can compute the duration of each leg by applying the first formula.
My first thought to apply the formula was to divide 30 km/h by 12 km, but that gives me 2.5 hours, which can't be right. I'm mixing up the numbers somehow.

To get duration, you are supposed to divide the change in distance by speed. Not speed by change in distance.

Remember, this is because speed is defined to be change in distance divided by duration.

For the first leg, would that be 12 km/ 30 km/h, yielding 0.4 hours?

Correct.

Wonderful. So the duration for the second leg of the trip would be approximately 0.47 hours. So then do I divide 52 km by 0.87 hours? That would yield an answer of approximately 59 km/h.

Well done!

Thank you! And thank you for helping! You may have just helped raised my current grade from a B+ to an A. 59 km/h is my final answer, correct?

The answer was more like 59.7 km/h, which you ought to round up to 60 km/h.

Thank you. I saw that on my calculator, but my teacher is stressing significant factors right now, and since each number was limited to two significant figures, I fear he may mark it as incorrect if I round to 60 or use three significant figures.

60 is two significant figures, and the only correct representation of your intermediate result 59.7... in two significant figures is 60. Not 59. 59 would be correct if you had, for example, 59.3...

Okay, thank you!