How Do You Calculate Velocity in Calculus-Based Physics?

AI Thread Summary
The discussion focuses on calculating the velocity of a 3kg box influenced by tension and gravity. The tension is described by the function T(t) = (36N/s)*t, leading to a derived acceleration formula a(t) = (2.2m/s^3)t - 9.8m/s^2. The user initially miscalculated the box's velocity at t = 1.0s and t = 3.0s, arriving at -8.7m/s instead of the correct -3.8m/s. Clarification on how to derive the acceleration from the tension function helped the user understand the solution better. The conversation highlights the importance of correctly applying Newton's second law in dynamic situations.
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Homework Statement

A 3kg box is held up high in the air by a a rope with negligible mass. The rope's tension is depends on the function of time T(t) = (36N/s)*t. The box is at rest at t = 0. Only the forces tension and gravity act on this box.

a) What is the velocity at t = 1.0s?
b) What is the velocity at t = 3.0s?

Homework Equations



I think that my acceleration formula is the problem, but I don't know why.

a(t) = (2.2m/s^3)t - 9.8m/s^2 I took out the mass to get the box's acceleration and then I just broke up the applied acceleration from the constant gravitational acceleration.

v(t) = (1.1m/s^3)t^2 - (9.8m/s^2)t This is my integral of the acceleration.

The Attempt at a Solution



a) a(1s) = 1.1m/s - 9.8m/s = -8.7m/s but the book's answer is -3.8m/s.

I just can't grasp how I am wrong.
 
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How did you get 2.2 m/s^3 from 36 N/s?
 
voko said:
How did you get 2.2 m/s^3 from 36 N/s?

ƩFy = T - mg = ma; a = 12m/s^2 - 9.8m/s^2 = 2.2m/s^2, so I divided out the mass to get the box's acceleration of 12m/s^2. Then, I ...

Lol, ok, in my attempt to answer your question I see how it can work I can get the right answer. Thanks for asking :)
 
You are welcome :)
 
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