How Do You Calculate Velocity in Calculus-Based Physics?

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student34
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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 :)