Tricky velocity/time problem without distance

In summary: Delta t = \Delta x + v_a In summary, a person takes a trip of 89.5 km/h, except for a 22-min rest stop, and spends an average of 77.8 km/h driving. They travel a total of 0.37 h during the trip.
  • #1
PhysicsBoyMan
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Homework Statement


A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22-min rest stop. If the person's average speed is 77.8 km/h
a) How much time is spent on the trip?
b) How far does the person travel?

Homework Equations


v1 = 89.5 km/h -> t
v2 = 0 km/h -> 0.37 h
AvgV = 77.8 km/h
a = 0 since speed is constant

The Attempt at a Solution


I have been given formulas throughout my textbook to solve for displacement, velocity, acceleration, and things like this. I will try to briefly show the kind of problems I run into and my thought process in solving this:

I'm looking for x (distance) and t (time) anywhere I can make it work basically.

average v = Δx / Δt I don't have Δx so I can't use this to find t.

v = vo + at I have original velocity but I can't isolate time because a=0.

All of my other equations don't work because they divide by 0, there are 2 unknown variables, or other such problems.

Im starting to wonder about how my "t"s are split into 2 parts in my question (0.37h and unknown) but in my equations there is only one variable t. I don't think my equations will work.

I can average the two velocities out to 89.5 km/h, but I can't figure out how to account for times which would make one velocity count for more than the other.

I tried to input 1h for my unknown t and calculate average velocity just to see if that would help me better understand the problem. I still don't know how to calculate average velocity without x (displacement). And I don't know how to find x with two different time values. I am really stuck!
 
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  • #2
PhysicsBoyMan said:

Homework Statement


A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22-min rest stop. If the person's average speed is 77.8 km/h
a) How much time is spent on the trip?
b) How far does the person travel?

Homework Equations


v1 = 89.5 km/h -> t
v2 = 0 km/h -> 0.37 h
AvgV = 77.8 km/h
a = 0 since speed is constant

average v = Δx / Δt I don't have Δx so I can't use this to find t.

The trip is made up of two legs, driving (1) and resting (2), combining the information from both legs is what you need to do to solve:

xtotal = x1 +x2

ttotal = t1 + t2 = t1 +0.37 h

express x1 in terms of t1 and V1 and then you can solve:

Vaverage = xtotal / ttotal
 
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  • #3
Hello @PhysicsBoyMan,

Welcome to Physics Forums! :smile:

PhysicsBoyMan said:

Homework Statement


A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22-min rest stop. If the person's average speed is 77.8 km/h
a) How much time is spent on the trip?
b) How far does the person travel?

Homework Equations


v1 = 89.5 km/h -> t
v2 = 0 km/h -> 0.37 h
AvgV = 77.8 km/h
a = 0 since speed is constant

The Attempt at a Solution


I have been given formulas throughout my textbook to solve for displacement, velocity, acceleration, and things like this. I will try to briefly show the kind of problems I run into and my thought process in solving this:

I'm looking for x (distance) and t (time) anywhere I can make it work basically.

average v = Δx / Δt I don't have Δx so I can't use this to find t.

You'll use that formula, so keep it handy. You'll actually use it twice. But before you do, I suggest defining your terms a little better (see below).

v = vo + at I have original velocity but I can't isolate time because a=0.

Yeah, don't worry about acceleration. It doesn't matter for this problem.

All of my other equations don't work because they divide by 0, there are 2 unknown variables, or other such problems.

Im starting to wonder about how my "t"s are split into 2 parts in my question (0.37h and unknown) but in my equations there is only one variable t. I don't think my equations will work.

Yes, you do need to split something apart. So as part of that split, start defining terms specifically. I suggest something along the lines of the following:

[itex] v: [/itex] velocity while driving.
[itex] v_a: [/itex] average velocity.
[itex] \Delta t: [/itex] driving time.
[itex] \Delta t_r: [/itex] resting time.
[itex] \Delta t_t: [/itex] total time.
[itex] \Delta x: [/itex] distance

Now see if you can simplify, and solve for the driving time, [itex] \Delta t [/itex], as an interim step. (Hint: You'll use your formula a couple of times in the process.)
 
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  • #4
billy_joule said:
The trip is made up of two legs, driving (1) and resting (2), combining the information from both legs is what you need to do to solve:

xtotal = x1 +x2

ttotal = t1 + t2 = t1 +0.37 h

express x1 in terms of t1 and V1 and then you can solve:

Vaverage = xtotal / ttotal

Thanks for the help. I expressed x1 in terms of t and v like you said. I have two occurences of t1. Is that normal? Is this ok so far? I tried to continue this path a few different ways but failed to come up with a single t value.

photo.jpg

upload a picture
 
  • #5
PhysicsBoyMan said:
Thanks for the help. I expressed x1 in terms of t and v like you said. I have two occurences of t1. Is that normal? Is this ok so far? I tried to continue this path a few different ways but failed to come up with a single t value.

What you have so far is correct.
If you multiply both sides by (t1 + 0.37 h) you'll get:

(77.8km/hr) (t1 + 0.37 h) = (89.5km/hr)(t1)

expand the LHS then solve for t1
 
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  • #6
I don't know how to isolate one occurrence of t1 when there is multiplication involved. Here are a few methods I tried and failed with. I can't imagine any other possible way.
photo.jpg

click image upload

Can someone please help me finish this it would be useful to learn how to isolate a variable when there are two of the same.
 
  • #7
PhysicsBoyMan said:
I don't know how to isolate one occurrence of t1 when there is multiplication involved. Here are a few methods I tried and failed with. I can't imagine any other possible way.Can someone please help me finish this it would be useful to learn how to isolate a variable when there are two of the same.
I don't understand your second line. What happened to the +0.37h?
Starting again from the first line, get all terms with a t1 factor on one side of the equation and all terms without it on the other.
 
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  • #8
haruspex said:
I don't understand your second line. What happened to the +0.37h?
Starting again from the first line, get all terms with a t1 factor on one side of the equation and all terms without it on the other.

What happened to the +0.37h in the second line is that Billy Joule said to expand the LHS and solve for t1, so I multiplied (77.8km/h) (t1+0.37h).

So back to the first line. I tried to get all terms with t1 factor on one side. This attempt is not shown in my picture, but I found that dividing both sides by (t1+0.37h) left me with a t1 divided by a t1 which equals 1 and I lose my t1. So besides dividing, how else can I get both occurrences of t1 to the same side?

In my second method in my picture (bottom blue box), I tried to take the whole right side (89.5km/h)(t1) and subtract that from both sides. That's why it is in square brackets and negative on the LHS. Then I tried to factor out t1 from that.

Then in lines 2 and 3 in the first blue box (I know my work is scattered all over the map) I tried to simply first to see if I would be able to gather t1 terms on the same side later. This did not work.

I tried to divide, I tried to subtract, but I was unable to get both t1 terms on the same side of the equation in the same numerator.
 
  • #9
I just had a bit of a breakthrough. If I can divide (km/h)(t1) / (km/h)(t1) then I would be left with a single t1 term. Only problem is that my t1 on the LHS is tied up in an addition equation. I tried to expand it by multiplying but someone said I did it the wrong way.
 
  • #10
[itex](77.8km/h)(t1 + 0.37H) ≠ (77.8km/h)(t1)(28.8km/h) [/itex]

Rather: [itex] x(y + z) = xy + xz[/itex]
 
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  • #11
So here is what I did. My answer sheet says that it should be 2.8h. This is the closest I got. I hope the discrepancy is just from leaving out a few decimal places here and there. Is this correct?

image.jpg

photo hosting
 
  • #12
Your solution looks about right to me. There is some error due to rounding (most notably involving the resting time), but that's not enough to explain the entire discrepancy. I think there might be a mistake in the answer sheet.
 
Last edited:
  • #13
collinsmark said:
Your solution looks about right to me. There is some error due to rounding (most notably involving the resting time), but that's not enough to explain the entire discrepancy. I think there might be a mistake in the answer sheet.
The question wording implies the 22 minutes rest is to be counted as part of the trip.
 
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  • #14
haruspex said:
The question wording implies the 22 minutes rest is to be counted as part of the trip.
Right. I'm just saying that 22/60 = 0.37 contains the bulk of the rounding errors. (Not enough to explain the discrepancy with the answer sheet though.)

[Edit: Oh, wait. I get it. Add in the 22/60 hours to the traveling time to get the total trip time. That comes out to around 2.8 hours, yes. Nevermind.]
 
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Related to Tricky velocity/time problem without distance

1. What is a tricky velocity/time problem without distance?

A tricky velocity/time problem without distance is a physics problem that involves calculating the velocity or time of an object's motion without knowing the distance it has traveled. This type of problem requires the use of equations and principles of motion to solve.

2. How do I approach solving a tricky velocity/time problem without distance?

To solve this type of problem, you should first identify the given information, such as the initial and final velocity, acceleration, and time. Then, you can use equations such as v = u + at and s = ut + 1/2at^2 to solve for the unknown variable.

3. Can I solve a tricky velocity/time problem without distance using a graph?

Yes, you can use a velocity-time graph to solve this type of problem. The slope of the line on the graph represents the acceleration, and the area under the line represents the displacement. By analyzing the graph, you can find the unknown variable.

4. Are there any shortcuts or tricks for solving tricky velocity/time problems without distance?

There are no shortcuts for solving these types of problems. However, it is helpful to have a strong understanding of the equations and principles of motion, as well as practice with similar problems to improve problem-solving skills.

5. What are some real-life examples of tricky velocity/time problems without distance?

Some real-life examples of these types of problems include calculating the time it takes for a car to reach a certain speed without knowing the distance it traveled, or determining the velocity of a ball thrown in the air without knowing the height it was thrown from.

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