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Alright no problem
use the above equation to find the displacement
So I would do d=0.5 x 5.0m/s^2? or somewhere else?
First one is the positive vector and negative for the second.
Alright let me walk you through
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.
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.
Notice that displacement only concerns the overall travelled distance from the initial position to your final position.
I hope this helps. Let me know where you lack comprehenion .
So do you subtract the first from the second then?
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?
What you got is an overall displacement.
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.
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
make sure to use negative acceleration too
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
@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
Physics always goes against our intuition.
d=(25m/s*5s)+(-5m/s^2)(5s)^2 Should get you the overall displacement. Initial velocity was calculated using at=v
Acceleration is change of velocity per unit of time so don't worry too much about always having to express it with seconds