An object's movement has a velocity given by v(t) = t^2-5t+5
A) What is the position function for the particle at any time t≥0?
For this section I calculated the anti derivative, which is \frac{1}{3}t^3-\frac{5}{2}t^2+5t+C (At least I think it is) And I'm not sure if that finishes the question or if I have to somehow calculate C... but I'm not sure how to do that.
B) When does the particle change directions?
C) Find the displacement and total distance traveled from time t = 0 to t = 5.
The anti derivative i got is 1/3t^3 - 5/2t^2 + 5t + C
1/3t^3 - 5/2t^2 + 5t + C is ok
let's call it s(t)
s(t) = 1/3t^3 - 5/2t^2 + 5t + C
you want the distance between t = 0 and t = 5
s(0) = 0-0+0 + C
s(5) = (1/3)(125) - (5/2)(25) + 5(5) + C
= 25/6 + C
distance covered between the two times = 25/6 + C - C = 25/6 <---- part c
b) it would change direction when v ' (t) = 0
v(t) = a(t) = 2t - 5 , where a(t) is the acceleration
2t - 5 = 0
t = 2.5
it would change direction when v(t) = 0 and v'(t) ≠ 0
That is, it stops and then reverses direction.
A) To find the position function for the particle at any time t≥0, you need to integrate the velocity function.
Given v(t) = t^2 - 5t + 5, the position function would be the antiderivative or integral of v(t) with respect to t.
∫v(t) dt = ∫(t^2 - 5t + 5) dt
Using the power rule for integration, we have:
∫t^2 dt - ∫5t dt + ∫5 dt
Integrating each term separately gives:
(1/3)t^3 - (5/2)t^2 + 5t + C
So, you were correct in calculating the antiderivative. However, you still need to find the value of the constant C.
To find C, you can use an initial condition. If the position of the object is known at a specific time, you can substitute that value into the position function and solve for C.
For example, if it is given that the position at time t = 0 is x = 0 (assuming x is the position), you can substitute these values into the position function:
0 = (1/3)(0)^3 - (5/2)(0)^2 + 5(0) + C
Simplifying this equation, you find:
0 = 0 + 0 + 0 + C
Therefore, C = 0.
So, the position function for the particle at any time t≥0 is:
x(t) = (1/3)t^3 - (5/2)t^2 + 5t
B) The particle changes its direction when the velocity changes from positive to negative or from negative to positive. In this case, we need to find the values of t for which the velocity function v(t) changes sign.
To find when v(t) = 0, we set the velocity function equal to zero and solve for t:
t^2 - 5t + 5 = 0
This quadratic equation can be solved using the quadratic formula:
t = [5 ± √(5^2 - 4(1)(5))] / 2
Simplifying, we have:
t = [5 ± √(25 - 20)] / 2
t = [5 ± √5] / 2
So, the particle changes direction at t = (5 - √5)/2 and t = (5 + √5)/2.
C) Displacement measures the change in position, while total distance traveled measures the overall distance covered.
To find the displacement from time t = 0 to t = 5, you can simply evaluate the position function x(t) at those points:
Displacement = x(5) - x(0)
Displacement = [(1/3)(5)^3 - (5/2)(5)^2 + 5(5)] - [(1/3)(0)^3 - (5/2)(0)^2 + 5(0)]
Simplifying this expression gives you the displacement from time t = 0 to t = 5.
To find the total distance traveled, you need to consider both positive and negative displacements. In this case, you can take the absolute value of the displacement.
Total Distance Traveled = |Displacement from t = 0 to t = 5|
Evaluate the expression for displacement and then take the absolute value to find the total distance traveled from time t = 0 to t = 5.