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.
x = 3 t^3 + 1
dx/dt = 9 t^2
d^2x/dt^2 = 18 t
so at t = 0 both acceleration and velocity are 0
velocity is always greater than 0 so the particle never backs up. Therefore all we need is the position at t = 5 minus the position at t = 1
x(5) = 251
x(1) = 7
------------
difference = 244
thank you!!!!!!!!!!! I don't think any of my teachers would deticate some of their time for us:D
a. To find the acceleration of the particle at t = 0, we need to find the second derivative of x(t) with respect to time (t).
The given equation for the position of the particle is x(t) = 2t^3 + 1.
To find the acceleration, we need to take the second derivative of x(t) with respect to t.
x'(t) = d(x(t))/dt = d/dt (2t^3 + 1) = 6t^2
Now, let's find the second derivative:
x''(t) = d^2(x(t))/dt^2 = d/dt (6t^2) = 12t
To find the acceleration at t = 0, substitute t = 0 into the equation for the second derivative:
x''(0) = 12(0) = 0
Therefore, 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 find the first derivative of x(t) with respect to time (t) and set it equal to 0.
The first derivative of x(t) is given by:
x'(t) = d(x(t))/dt = d/dt (2t^3 + 1) = 6t^2
To find when the acceleration is 0, we set the second derivative equal to 0:
x''(t) = 12t = 0
Solving for t, we find t = 0.
Now, substitute t = 0 into the equation for velocity:
x'(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 interval [0, 5].
The velocity function was found in part b to be x'(t) = 6t^2.
To find the total distance, we integrate the absolute value of the velocity function:
∫[0,5] |x'(t)| dt = ∫[0,5] |6t^2| dt
Since the absolute value function can be split into two cases, we have:
∫[0,5] 6t^2 dt + ∫[0,5] -6t^2 dt
Integrating each term separately, we get:
= 2t^3 |[0,5] - 2t^3 |[0,5]
= 2(5)^3 - 2(0)^3 + 2(0)^3 - 2(0)^3
= 250
Therefore, the total distance traveled by the particle from t = 0 to t = 5 is 250 units.