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please help! phys ... 2D motion

 
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Sep19-05, 12:06 AM   #1
 

please help! phys ... 2D motion


At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.

i am asked to find the acceleration at time t.
i know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.
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Sep19-05, 12:52 AM   #2
 
... ??
Sep19-05, 12:57 AM   #3
 
Quote by lostinphys
At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.

i am asked to find the acceleration at time t.
i know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.

I think you're on the right track, but it looks like you're trying to use time as a vector here. remember, it divides both components of your [tex]\Delta v[/tex] vector.
Sep19-05, 12:58 AM   #4

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please help! phys ... 2D motion


Well the first part, 6.2i + 8.1j is correct. You divide by 2.20 seconds but it looks more like (6.2/2.20)i + (6.10/2.20)j with m/s^2 being the units for the vector. You can treat them seperately and think of it like 6.2/2.20 as the dv/dt in the i direction for example.
Sep19-05, 01:10 AM   #5
 
The vector representing acceleration would then be (6.2/t)i + (8.1/t)j ?
Sep19-05, 01:11 AM   #6
 
That sounds right.
Sep19-05, 01:23 AM   #7
 
thanks !!
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