What is the Average Speed and Acceleration of a Car on a Straight Road?

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Homework Help Overview

The discussion revolves around a problem involving a car accelerating from rest to 45 miles per hour over a 5-second interval on a straight road. Participants are tasked with calculating average speed, acceleration, and instantaneous speed at specific times during the acceleration period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of average speed and acceleration, questioning the correct interpretation of time intervals and units. Some express uncertainty about how to derive instantaneous speed at specific times and the relationship between average and instantaneous speed.

Discussion Status

There is ongoing exploration of the relationships between average speed, instantaneous speed, and acceleration. Some participants have provided guidance on using equations to relate these quantities, while others are clarifying the implications of uniform acceleration. Multiple interpretations of the problem are being discussed, particularly regarding the correct application of time and units.

Contextual Notes

Participants note potential confusion regarding the time values used in calculations and the units of acceleration. There is an emphasis on ensuring clarity in the definitions of speed and velocity, as well as the implications of uniform acceleration on the results.

  • #31
It would be the same?
 
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  • #32
Tonia said:
Would the expression for speed be: v = m = (45 mi/hr - 0 mi/hr)/2.5 sec. = 18 mi/hr/sec.??
No, that's calculating an acceleration, not a speed.
Write out the general equation that relates speed, time and (uniform) acceleration.
You have a value for the acceleration, and a value for the speed, and we're leaving the time as an unknown t. Plug those into the equation.
 
  • #33
Tonia said:
It would be the same?
What would be the same as what?

Edit: Ok, you were replying to this:
Right, so how would the average height (=average speed) compare with the height half way along (=instantaneous speed at half time)?

Yes.
(Please use the Reply button or the quote button so people know what you are answering.)
 
  • #34
The equation would be Vf = Vo + at??
 
  • #35
Tonia said:
The equation would be Vf = Vo + at??
Right, so plug in the values for v0, vf and a that are appropriate to question d).
 
  • #36
45 mi/hr = 0 + 9 mi/hr/sec times t
Solve for t??
 
  • #37
t would be 5.
 
  • #38
Tonia said:
45 mi/hr = 0 + 9 mi/hr/sec times t
Solve for t??
What was the speed you calculated in part a)?
 
  • #39
22.5 mi/hr
 
  • #40
Tonia said:
22.5 mi/hr
Right. That is the speed you are told to use in answering part d).
 
  • #41
so t is 2.5 seconds and that's the answer to d.
 
  • #42
Tonia said:
so t is 2.5 seconds and that's the answer to d.
Yes.
 
  • #43
Okay. thanks for your help!
 
  • #44
Let me ask you one more question just to make sure: for problems a) and c), the answer is the same? 22.5mi/hr??
 
  • #45
Tonia said:
Let me ask you one more question just to make sure: for problems a) and c), the answer is the same? 22.5mi/hr??
Yes.
 
  • #46
Okay, thanks!
 

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