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Time when car stopped given initial velocity and acceleration

  • Thread starter 1MileCrash
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  • #1
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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.
 

Answers and Replies

  • #2
lightgrav
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nothing wrong here. is there a part 2 of the question that you didn't answer?
 
  • #3
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No, and even if there was, the system allows partial submissions.
 
  • #4
lightgrav
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they goofed up. it happens.
 
  • #5
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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.)
 
  • #6
lewando
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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.
 
  • #7
lightgrav
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maybe they do have changing acceleration (didn't notice the "t", sorry)

for problem 2 they want _velocity_ , not distance ...
 
  • #8
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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?
 
  • #9
lightgrav
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I don't know . . . you're ignoring units! where did the 6 t come from ? a = 3 m/s^2
 
  • #10
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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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