Velocity to acceleration then conversion from km/h to m/s2 confusion

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Homework Help Overview

The discussion revolves around a physics problem involving a car traveling at a constant speed of 60 km/h for 5.3 hours, with a focus on understanding acceleration in this context. Participants explore the definitions and implications of constant speed and acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of acceleration and its relationship to constant velocity. Questions arise about the initial and final speeds, and whether the speed changes during the time interval. Some participants attempt to calculate acceleration based on their interpretations of the problem.

Discussion Status

There is a productive exploration of the concept of constant speed leading to zero acceleration. Participants are questioning assumptions about the initial conditions and the nature of the motion described in the problem. Some have reached an understanding that constant velocity implies no change in speed, thus resulting in zero acceleration.

Contextual Notes

Participants note the importance of definitions and the implications of the term "constant" in the context of speed and acceleration. There is an acknowledgment of the need for clarity regarding initial and final speeds in the problem setup.

survivorboiii
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A car travels in a straight line for 5.3 h at a constant speed of 60 km/h.
What is its acceleration? Answer in units of m/s2

So far, to my understanding so far acceleration is change in velocity over time. So 60 over 5.3, which is 11.32 km/h2

Then I converted to m first, which is 1132 m/h2, then from h2 to sec, which is .00524 m/sec2.

But that isn't the answer, what did I do wrong?
 
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"a constant speed of 60 km/h.
What is its acceleration?"

What does CONSTANT speed mean, when you are traveling along a straight line? In particular, what is the acceleration then?
 
I'm not understanding what your asking. The acceleration is (60-0)/time right?
 
What was the INITIAL speed?
What is the FINAL speed?

In particular, in those 5.3 hours, does the speed change?
 
hi survivorboiii! welcome to pf! :smile:
survivorboiii said:
… acceleration is change in velocity over time. …

hint: what does "change" mean? :wink:
 
The car probably started at 0km/h, then speeds up to 60km/h where it stayed for 5.3hr
 
"0km/h, then speeds up to 60km/h "
Really?
What does the book say about the speed during the time interval we're looking at?
 
Enlighten me, I don't have my book with me. I guess I'm still confused about speed, vel, and acc
 
What does the word CONSTANT mean?
 
  • #10
constant velocity means acceleration is 0 doesn't it because the velocity isn't changing
 
  • #11
survivorboiii said:
constant velocity means acceleration is 0 doesn't it because the velocity isn't changing

Correct!

So what was wrong with your two statements:

1. " The car probably started at 0km/h, then speeds up to 60km/h where it stayed for 5.3hr"

2. "The acceleration is (60-0)/time right? "

In particular:
How should you reframe the equation in 2. ?
 
  • #12
Would it be 0-60 over time?
 
  • #13
Why?
You said yourself:
"constant velocity means acceleration is 0 doesn't it because the velocity isn't changing "

How can the speed change from 0 to 60 if it isn't changing?
 
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  • #14
(just got up :zzz:)

average acceleration = final velocity minus initial velocity divided by time

instantaneous acceleration (usually just called "acceleration") = final velocity after an extremely short time minus initial velocity divided by that extremely short time = the slope of the velocity-time graph :smile:

this question is asking for acceleration​

so draw a graph … what is the slope? :wink:
 
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  • #15
IT just hit me...the acceleration is 0 because the vel is constant! :) Thanks both of you!
 
  • #16
survivorboiii said:
IT just hit me...the acceleration is 0 because the vel is constant! :) Thanks both of you!

:smile:
PS:
Remember to give your answer as 0m/s^2, because the exercise specifically asked you to include units
 

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