Using the equation x(t)= t^3 - 12t + 1, find the displacement during the first 3 seconds. And then find the velocity during the first 3 seconds.
See, I know how to find velocity, but there's a question after that that says to find velocity when t=3...so how do you find it DURING the 3 seconds?
I can only think they mean the average velocity during the 1st 3 seconds.
So, get the displacement and divide by the time for average velocity over the interval.
To find the displacement during the first 3 seconds, we need to evaluate the equation x(t) = t^3 - 12t + 1 at both t = 0 and t = 3.
For t = 0:
x(0) = (0)^3 - 12(0) + 1 = 0 - 0 + 1 = 1
For t = 3:
x(3) = (3)^3 - 12(3) + 1 = 27 - 36 + 1 = -8
The displacement during the first 3 seconds can be calculated by finding the difference between the positions at these two times:
Displacement = x(3) - x(0) = -8 - 1 = -9
To find the velocity during the first 3 seconds, we can differentiate the equation x(t) = t^3 - 12t + 1 with respect to time (t) to obtain the equation for velocity, v(t).
v(t) = d/dt (t^3 - 12t + 1)
= 3t^2 - 12
Now we can evaluate v(t) at t = 0 and t = 3 to find the velocities during those times:
For t = 0:
v(0) = 3(0)^2 - 12 = -12
For t = 3:
v(3) = 3(3)^2 - 12 = 27 - 12 = 15
Therefore, the velocity during the first 3 seconds is -12 m/s when t = 0 and 15 m/s when t = 3.
To find the displacement during the first 3 seconds, we can substitute t = 0 and t = 3 seconds into the equation x(t) = t^3 - 12t + 1.
1. Substituting t = 0:
x(0) = (0)^3 - 12(0) + 1
= 0 - 0 + 1
= 1
The displacement at t = 0 is 1.
2. Substituting t = 3:
x(3) = (3)^3 - 12(3) + 1
= 27 - 36 + 1
= -8
The displacement at t = 3 seconds is -8.
So, during the first 3 seconds, the displacement changes from 1 to -8.
Now, let's find the velocity during the first 3 seconds. The velocity can be calculated by taking the derivative of the displacement function x(t).
Given the equation: x(t) = t^3 - 12t + 1
To find the velocity, we differentiate x(t) with respect to time t:
v(t) = d(x(t))/dt
Differentiating each term of the equation, we get:
v(t) = d(t^3)/dt - d(12t)/dt + d(1)/dt
= 3t^2 - 12
Substituting t = 0 and t = 3 into the velocity equation:
1. Substituting t = 0:
v(0) = 3(0)^2 - 12
= 0 - 12
= -12
The velocity at t = 0 is -12.
2. Substituting t = 3:
v(3) = 3(3)^2 - 12
= 3(9) - 12
= 27 - 12
= 15
The velocity at t = 3 seconds is 15.
So, during the first 3 seconds, the velocity changes from -12 to 15.