A particle moves on the x –axis so that its position at any time is given by x(t) = 2t3 + 1.
a. Find the acceleration of the particle at t = 0.
b. Find the velocity of the particle when its acceleration is 0.
c. Find the total distance traveled by the particle from t = 0 to t = 5.
To find the answers to these questions, we need to differentiate the position function to get the velocity and acceleration functions. Let's start step by step:
a. To find the acceleration of the particle at t = 0, we need to differentiate the position function twice with respect to time.
x(t) = 2t^3 + 1
Differentiating once, we get the velocity function:
v(t) = d/dt (2t^3 + 1)
= 6t^2
Differentiating again, we get the acceleration function:
a(t) = d/dt (6t^2)
= 12t
Now, substitute t = 0 into the acceleration function to find the acceleration at t = 0:
a(0) = 12(0)
= 0
So, the acceleration of the particle at t = 0 is 0.
b. To find the velocity of the particle when its acceleration is 0, we need to set the acceleration function equal to 0 and solve for t.
a(t) = 12t = 0
Here, we see that the acceleration is 0 when t = 0. So, we need to find the velocity at t = 0.
v(t) = 6t^2
Now, substitute t = 0 into the velocity function to find the velocity when the acceleration is 0:
v(0) = 6(0)^2
= 0
Therefore, the velocity of the particle when its acceleration is 0 is 0.
c. To find the total distance traveled by the particle from t = 0 to t = 5, we need to integrate the absolute value of the velocity function over the time interval [0, 5].
v(t) = 6t^2
To find the total distance traveled, we need to integrate the absolute value of the velocity function over the interval [0, 5]:
Total distance = ∫[0, 5] |v(t)| dt
∫[0, 5] |6t^2| dt = ∫[0, 5] 6t^2 dt
Integrating, we get:
= [2t^3] evaluated from 0 to 5
= (2(5^3) - 2(0^3))
= (2(125) - 0)
= 250 units
Therefore, the total distance traveled by the particle from t = 0 to t = 5 is 250 units.