What is the velocity and position of a bus after a given time?

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SUMMARY

The discussion focuses on calculating the velocity and position of a bus given its acceleration function a(t) = 1.13t m/s³. For Part A, the correct approach to find the bus's velocity at time t_2 = 2.15 s, starting from an initial velocity of 4.93 m/s at t_1 = 1.01 s, involves using the equation Vf = Vo + a(tf - to) with the appropriate time difference. For Part B, to determine the position at t_2, a different kinematic equation must be applied, emphasizing the need to account for the time interval when calculating changes in position.

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  • Understanding of kinematic equations, specifically Vf = Vo + a(tf - to)
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Homework Statement


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?

Homework Equations



Vf = Vo + at

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