Calculating Time and Acceleration in Constantly Accelerating Cars

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SUMMARY

The discussion focuses on calculating time and acceleration for a car moving at constant acceleration, specifically between speeds of 40 km/h and 80 km/h over a distance of 250 meters. The correct approach involves using the average velocity, which is determined to be 60 km/h (16.67 m/s), leading to a travel time of 15 seconds. The acceleration is calculated using the change in velocity over time, resulting in an acceleration of approximately 0.74 m/s². Participants emphasized the importance of using the correct kinematic equations and understanding the distinction between average velocity and instantaneous velocity.

PREREQUISITES
  • Understanding of 1-D kinematics
  • Familiarity with the equations of motion under constant acceleration
  • Basic knowledge of unit conversions (e.g., km/h to m/s)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the equations of motion for constant acceleration, specifically vf² = vi² + 2aΔx
  • Learn how to convert units between kilometers per hour and meters per second
  • Explore the concept of average velocity and its application in kinematic problems
  • Review the principles of calculus-based physics as they apply to motion
USEFUL FOR

Students in high school or introductory college physics courses, particularly those studying kinematics and seeking to improve their problem-solving skills in physics. This discussion is also beneficial for educators looking for examples of common student misconceptions in physics calculations.

Manh
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Members are reminded to use the provided Formatting Template for homework posts!
In a car moving at constant acceleration, you travel 250 m between the instants at which the speedometer reads 40 km/h and 80 km/h.
How many seconds does it take you to travel the 250 m?
What is your acceleration?

I have thought of finding the average of 40 km/h and 80 km/h to start the problem but I'm unsure about that.
 
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You should use the homework template, to begin with.

There are some basic equations that describe objects under constant acceleration. Have you tried applying them to this situation?
 
I think the one I found does not apply to this situation. The correct equation may be a = v/t.
 
You don't know neither the acceleration nor the time interval,so you are going to need one more equation,
 
I have not received any other equation in class and therefore I give it a try by looking up equations around the internet. I just wonder if I'm supposed to find the average speed before I start to do anything else.
 
Manh said:
I think the one I found does not apply to this situation. The correct equation may be a = v/t.

Is this a high school class? There should be a textbook or some sort of reference material you have for the course.

The link you posted previously are all special cases of the formulas I was referring to, specifically arranged and simplified for 'free-fall' constant acceleration. The problem is that your object is not under constant 'free-fall' acceleration.
 
I think I might put in it in the wrong section. It is a problem from calculus-based physics on MasteringPhysics and I don't receive any further information beside it.
 
This is the correct section. You don't need to resort to using calculus on such a problem.

Manh said:
I have not received any other equation in class and therefore I give it a try by looking up equations around the internet. I just wonder if I'm supposed to find the average speed before I start to do anything else.
No, you don't need to find the average velocity, that won't help.

There is a sticky thread in the forum you posted in with introductory physics formulae:
https://www.physicsforums.com/threads/introductory-physics-formulary.110015/

You have an initial velocity(v0 ), a final velocity (v) and a displacement (Δx), you want to find the acceleration (a).
Try find the equation that has those.
 
  • #10
Manh said:
I think I might put in it in the wrong section. It is a problem from calculus-based physics on MasteringPhysics and I don't receive any further information beside it.

Do you have access to an e-text through your MasteringPhysics? You need to know the equations of 1-D kinematics to help you solve this problem. I highly doubt that many students will get very far through a calculus based physics course without having some kind of standard reference material to help them, especially if your teacher/professor is not actually providing you with these equations in class in a straightforward way.
 
  • #11
I found the equations from this link: http://neutrino.otterbein.edu/~tagg/Courses/Current/Phys1500/equation_sheet.pdf
Should I find delta v to apply it to the equation that has t?
 
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  • #12
billy_joule said:
No, you don't need to find the average velocity, that won't help.

Sorry, I was wrong here. Finding Vaverage will work to find t, the rest of post 9 will help you find a.

Manh said:
I found the equations from this link: http://neutrino.otterbein.edu/~tagg/Courses/Current/Phys1500/equation_sheet.pdf
Should I find delta v to apply it to the equation that has t?
Which one? 4 out of 5 of those equations have t in them...

It's best you make some attempt and we can check it.
 
Last edited by a moderator:
  • #13
Should I consider of delta v instead of average v?
 
  • #14
I see this equation may be helpful: xf = xi + v * delta t.

According to the equation I got approximately 22.50 s after finding delta v and converting all the units.
 
  • #15
That is not correct.

Show you're working (with units) and we'll see where you went wrong.
 
  • #16
1) delta v = 40 km/h = 11.11 m/s
2) 250 m = 0 m + 11.11 m/s * delta t
3) delta t = 250 m / 11.11 m/s
4) delta t = 22.50 s
 
  • #17
Manh said:
1) delta v = 40 km/h = 11.11 m/s
2) 250 m = 0 m + 11.11 m/s * delta t
3) delta t = 250 m / 11.11 m/s
4) delta t = 22.50 s
You take velocity to be constant and equal to 40 km/h.The problem states there is acceleration and the velocity goes up to 80 km/h.
Edit:Sorry,you obviously don't.It's late and my eyes betrayed me.
 
  • #18
But it also states constant acceleration at the beginning of the problem. I'm very confused!
 
  • #19
Manh said:
But it also states constant acceleration at the beginning of the problem. I'm very confused!

Yes, the car accelerates at a constant rate from 40 to 80 km/hr.

22.5 seconds is the time it take to travel 250 metres traveling at 40 km/hr - Not what you are looking for.

What is the average velocity of the car?
How long does it take to travel 250 metres at that average velocity?
 
  • #20
Yes,the problem is you have calculated the average velocity the wrong way.
 
  • #21
billy_joule said:
What is the average velocity of the car?
How long does it take to travel 250 metres at that average velocity?
v average = 60 km/h = 16.67 m/s
t = 15 s

Does it look correct now?
 
  • #22
Yes,well done.
 
  • #23
And a = 16.67 m/s / 15 s = 1.11 m/s^2?
 
  • #24
Manh said:
And a = 16.67 m/s / 15 s = 1.11 m/s^2?
No.
It's best to think carefully what it is you have and what it is you want to find before calculating anything.

Acceleration is the rate of change of velocity.
What is the change of velocity in this case?
 
  • #25
Can I say a = 0 m/s^2 because constant acceleration is mentioned in the problem?
 
  • #26
To change velocity, acceleration must occur. You've already found that change in velocity on page one.

Constant does not mean zero, it means it doesn't change with time. That is, the acceleration is the same for the entire 15 seconds.
 
  • #27
vf^2 = vi^2 + 2a(delta x)
22.22^2 = 11.11^2 + 2a(250)

Is this the right way to find a?
 
  • #28
The equation you used in post #23 is fine. Though, that equation (in #27) will work it is unnecessarily complicated.

The problem is that you used 16.67 m/s in post #23, that is the average velocity, you need to use the change in velocity.

Acceleration = Change in velocity / change in time
 
  • #29
a = 11.11 m/s / 15 s
a = 0.74 m/s^2 ?
 
  • #30
This would be it.Also,nice work on not forgetting to put the units of measurement after your results.
 

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