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alyssajobug
 one year ago
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?
alyssajobug
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0use the above equation to find the displacement

alyssajobug
 one year ago
Best ResponseYou've already chosen the best response.0So I would do d=0.5 x 5.0m/s^2? or somewhere else?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First one is the positive vector and negative for the second.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright let me walk you through

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0Notice that displacement only concerns the overall travelled distance from the initial position to your final position.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I hope this helps. Let me know where you lack comprehenion .

alyssajobug
 one year ago
Best ResponseYou've already chosen the best response.0So do you subtract the first from the second then?

alyssajobug
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0What you got is an overall displacement.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0@peachp Thanks for your correction I totally forgot to account for the preexisting velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0make sure to use negative acceleration too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0Physics always goes against our intuition.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0d=(25m/s*5s)+(5m/s^2)(5s)^2 Should get you the overall displacement. Initial velocity was calculated using at=v

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Acceleration 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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