- #1
StrandedStudent
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Stressed first year university student here, fresh out of high school. I took physics in both grade 11 and 12, and thought I had a pretty good grasp on it; that is until this week. Introduction to derivatives and integrals to get from x(t) to v(t) to a(t) and vice-versa. I have a pretty good grasp on how to obtain the derivatives and how to do the integral, I just have no clue on how to solve for average velocities (with or without constant accelerations??)
The attached file is the assignment that I am stuck on, but may have a slight understanding of how to attempt them.
My thinking:
1.a) To go from v(t) to x(t), I simply take the integral of v(t). v(t)= 5t^2 + 3t + 2 m/s
Integral of v(t)= 5/3t^3 + 3/2t^2 + 2t + constant = x(t)
Is there a way to get the constant without any other information?
x(t)= 5/3t^3 + 3/2t + 2t + constant m
b) To go from v(t) to a(t), I take the derivative.
Derivative of v(t)= 10t + 3 = a(t)
a(t)= 10t + 3 m/s^2
c) Completely lost here. Am I able to just plug in the two values of t and divide by 2? My professor said something about only being able to do this if the acceleration is constant, which it's not (is it?) due to the variable t being in the equation for acceleration.
2. My thinking is that I could take the derivative of x(t) to get v(t) and a(t) and plug in the given values for when t=0. Is this correct?
Any help would be greatly appreciated
The attached file is the assignment that I am stuck on, but may have a slight understanding of how to attempt them.
My thinking:
1.a) To go from v(t) to x(t), I simply take the integral of v(t). v(t)= 5t^2 + 3t + 2 m/s
Integral of v(t)= 5/3t^3 + 3/2t^2 + 2t + constant = x(t)
Is there a way to get the constant without any other information?
x(t)= 5/3t^3 + 3/2t + 2t + constant m
b) To go from v(t) to a(t), I take the derivative.
Derivative of v(t)= 10t + 3 = a(t)
a(t)= 10t + 3 m/s^2
c) Completely lost here. Am I able to just plug in the two values of t and divide by 2? My professor said something about only being able to do this if the acceleration is constant, which it's not (is it?) due to the variable t being in the equation for acceleration.
2. My thinking is that I could take the derivative of x(t) to get v(t) and a(t) and plug in the given values for when t=0. Is this correct?
Any help would be greatly appreciated