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Acceleration of a Busby CursedAntagonis
Tags: acceleration 
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#1
Sep2307, 11:05 PM

P: 23

1. The problem statement, all variables and given/known data
The acceleration of a bus is given by a (t)= alpha t, where alpha = 1.13 m/s^3 is a constant. Part A) If the bus's velocity at time t_1 = 1.01 s is 4.93 m/s, what is its velocity at time t_2 = 2.15 s? Part B) If the bus's position at time t_1 = 1.01 s is 6.01 m, what is its position at time t_2 = 2.15 s? 2. Relevant equations Vf = Vo + at 3. The attempt at a solution I am stuck on Part A and Part B. I am applying the Vf = Vo + at equation. I set it up as Vf = 4.93 + (alpha * t)(2.15s) and the answer that I get is wrong. Anyone have any idea how to approach this? Thanks. 


#2
Sep2407, 04:19 AM

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P: 9,781

Your equation that you are using for part (A) is correct, perhaps if you showed us your numbers we could point out whats going wrong. However, note that part (B) us asking for position, so you need a different kinematic equation.



#3
Sep2407, 09:44 AM

P: 2

Remember to take the difference of the two times when using v(1.01) as Vo in the equation. It is a very common mistake.
The real equation: Vf = Vo + a(tfto) for constant a. 


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