Time when car stopped given initial velocity and acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving a car's motion, specifically its initial velocity and acceleration, and the time it takes to come to a stop. The subject area includes kinematics and the analysis of motion under variable acceleration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between initial velocity and time to stop under a non-constant acceleration. There are attempts to derive equations for velocity and questions about the nature of acceleration being variable versus constant.

Discussion Status

The discussion is active, with participants questioning the setup of the problems and the implications of the acceleration being a function of time. Some have provided insights into the calculations and interpretations, while others are clarifying the requirements of the problems.

Contextual Notes

There is mention of partial submissions allowed by the system, and participants are navigating the implications of changing acceleration in their calculations. The original poster expresses confusion regarding the correctness of their approach and the interpretation of the problem statements.

1MileCrash
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Homework Statement



At time t = 0, a car has a velocity of 16 m/s. It slows down w/an acceleration given by -0.50t in m/s^2 for t in seconds. It stops at t =?

Homework Equations





The Attempt at a Solution



v = 16 - 0.5t
16/.5 = t
16 x 2 = t
32 = t

Did I set this up wrong? My answer is incorrect. It even works intuitively, if speed is reduced by .5 every second, starting at 16, it will take 32 subtractions of .5 to equal 0.
 
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nothing wrong here. is there a part 2 of the question that you didn't answer?
 
No, and even if there was, the system allows partial submissions.
 
they goofed up. it happens.
 
Okay. Just to make sure they aren't 2 for 2..
"A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s2 south. After 6 seconds its velocity will be:"

I say 11 m/s south, but this is no option.

x = 25t - 3t^2
v = 25 - 6t
v(6) = 25 - 36
v = -11 (or 11 south, as the option format it.)
 
It slows down w/an acceleration given by -0.50t in m/s^2 for t in seconds.

Could this mean the acceleration is not constant? -0.50t m/s2 is different from -0.50 m/s2.
 
maybe they do have changing acceleration (didn't notice the "t", sorry)

for problem 2 they want _velocity_ , not distance ...
 
lewando said:
Could this mean the acceleration is not constant? -0.50t m/s2 is different from -0.50 m/s2.

I'll try

for problem 2 they want _velocity_ , not distance ...

How is what I found distance?
 
I don't know . . . you're ignoring units! where did the 6 t come from ? a = 3 m/s^2
 
  • #10
Guy, y'all were right. Acceleration was not constant. I defined acceleration with a = -.5t + 16 and integrated for velocity function, for a result of 8 seconds when velocity is zero which is correct.

I don't know . . . you're ignoring units! where did the 6 t come from ? a = 3 m/s^2

I differentiated, but I see the error. My position function should have 1.5 as the second coefficient, and the derivative should be back at 3.

So, they didn't goof up at all.
 
Last edited:

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