alyssajobug
  • alyssajobug
Please help, I'm not sure how to even approach this question... A Porsche accelerates from a stop light at 5.0m/s^2 for 5s, and then decelerates at 5.0m/s^2 for 3s. How far has it traveled?
Physics
katieb
  • katieb
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anonymous
  • anonymous
Alright no problem
anonymous
  • anonymous
d=0.5at^2
anonymous
  • anonymous
use the above equation to find the displacement

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alyssajobug
  • alyssajobug
So I would do d=0.5 x 5.0m/s^2? or somewhere else?
anonymous
  • anonymous
First one is the positive vector and negative for the second.
anonymous
  • anonymous
Alright let me walk you through
anonymous
  • anonymous
For the first displacement you can calcaulate the magnitude of displacement using 0 d=0.5(5m/s)(5s)^2 Notice how the units of seconds cancel out and what's left is "m" which accounts for the displacement.
anonymous
  • anonymous
You do that for the first half of the question and then a=0.5(5m/s^2)(3m)^2 Subtract the first with the second to get the overall displacement.
anonymous
  • anonymous
Notice that displacement only concerns the overall travelled distance from the initial position to your final position.
anonymous
  • anonymous
I hope this helps. Let me know where you lack comprehenion .
alyssajobug
  • alyssajobug
So do you subtract the first from the second then?
alyssajobug
  • alyssajobug
So when solving for this I take the 7.5 that I got from the second equation and subtract it from the 12.5 I got from the first equation to get 5m^2 right? or did I misunderstand? Robert136?
anonymous
  • anonymous
What you got is an overall displacement.
anonymous
  • anonymous
For the first accelerating part you can use that equation because it's starting from rest. I don't think you can use it for decelerating part because it has velocity.
anonymous
  • anonymous
You can use v = v0 + at to find the velocity at the end of the acceleration. Then use d = 0.5at² + vt to find the displacement for that part
anonymous
  • anonymous
@peachp Thanks for your correction I totally forgot to account for the preexisting velocity
anonymous
  • anonymous
make sure to use negative acceleration too
anonymous
  • anonymous
no problem I meant use d = 0.5at² + vt to find the displacement for the deceleration part. Where a will be -5. And then you'd add the 2 displacements to get the overall
anonymous
  • anonymous
@peachpi Truely so. It would have been wiser first calculate the initial velocity and then account for the negative acceleration. However the counterintuitive question is whether to fully consider the initial velocity as being part of the whole but as far as the equation goes negative acceleration looks after it all
anonymous
  • anonymous
Physics always goes against our intuition.
anonymous
  • anonymous
d=(25m/s*5s)+(-5m/s^2)(5s)^2 Should get you the overall displacement. Initial velocity was calculated using at=v
anonymous
  • anonymous
Acceleration is change of velocity per unit of time so don't worry too much about always having to express it with seconds

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