Speed/Velocity Problem

  • Thread starter Medgirl314
  • Start date
  • #1
Medgirl314
561
2

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.

Thanks in advance!
 

Answers and Replies

  • #2
voko
6,054
391
What do you need to know to find the average speed?
 
  • #3
Medgirl314
561
2
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.
 
  • #4
voko
6,054
391
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}}}

$$
 
  • #5
Medgirl314
561
2
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.

Thanks for replying!
 
  • #6
Medgirl314
561
2
*with only t instead of delta t. Sorry. Auto-correct is out to get me.
 
  • #7
voko
6,054
391
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?
 
  • #8
Medgirl314
561
2
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?
 
  • #9
voko
6,054
391
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?
 
  • #10
Medgirl314
561
2
Would delta x=52 km, and delta t=115 km/h ? Or vice-versa?
 
  • #11
voko
6,054
391
## \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.
 
  • #12
Medgirl314
561
2
Great, thank you! So how do I figure out the time if I'm only given speed and distance?
 
  • #13
Kot
57
1
Time can be found by dividing distance by speed.
 
  • #14
voko
6,054
391
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.
 
  • #15
Medgirl314
561
2
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.
 
  • #16
voko
6,054
391
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.
 
  • #17
Medgirl314
561
2
For the first leg, would that be 12 km/ 30 km/h, yielding 0.4 hours?
 
  • #18
voko
6,054
391
Correct.
 
  • #19
Medgirl314
561
2
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.
 
  • #20
voko
6,054
391
Well done!
 
  • #21
Medgirl314
561
2
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?
 
  • #22
voko
6,054
391
The answer was more like 59.7 km/h, which you ought to round up to 60 km/h.
 
  • #23
Medgirl314
561
2
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.
 
  • #24
voko
6,054
391
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...
 
  • #25
Medgirl314
561
2
Okay, thank you!
 
  • #26
voko
6,054
391
I do have to say that 60 may also imply it has only one significant figure. Unless you have been told how to indicate symbolically that 60 is correct to two significant figures, just write that to avoid any confusion.
 
  • #27
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,809
1,670
In problems like these, pay attention to the units, which can help to eliminate confusion on how to apply the formulas.
 
  • #28
Medgirl314
561
2
@Voko, Right! I forgot that 0 wasn't a significant figure after other numbers. Nearly midnight is clearly not the time to be doing physics.

Thanks, @Steam King.
 

Suggested for: Speed/Velocity Problem

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
1
Views
933
Replies
24
Views
15K
Replies
11
Views
12K
Replies
4
Views
15K
Replies
1
Views
6K
Replies
12
Views
9K
Replies
3
Views
2K
Replies
2
Views
9K
Top