Consider a particle moving along the x-axis where
x(t)
is the position of the particle at time t,
x' (t)
is its velocity, and
x'' (t)
is its acceleration.
x(t) = t3 − 12t2 + 21t − 7, 0 ≤ t ≤ 10
(a) Find the velocity and acceleration of the particle.
x' (t) =
x'' (t) =
no, duh. it is negative between the roots, which are NOT at 0 and 1!
Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)
moving to the right means x is increasing, so x' > 0
x = t^3 − 12t^2 + 21t − 7
x' = 3t^2 - 24t + 21 = 3(t^2-8t+7) = 3(t-1)(t-7)
x" = 6t - 24 = 6(t-4)
so, check where x' > 0 (hint: not between the roots, since it is a parabola which opens up)
Remember the domain.
(0,1)?
To find the velocity and acceleration of the particle, we need to take the derivatives of the given position function x(t) = t^3 - 12t^2 + 21t - 7.
First, let's find the derivative of x(t) to get the velocity function x'(t):
x'(t) = d/dt(t^3 - 12t^2 + 21t - 7)
To take the derivative, we use the power rule and the sum/difference rule of derivatives. Applying these rules, we get:
x'(t) = 3t^2 - 24t + 21
So, the velocity of the particle is given by the function x'(t) = 3t^2 - 24t + 21.
Next, let's find the derivative of x'(t) to get the acceleration function x''(t):
x''(t) = d/dt(3t^2 - 24t + 21)
Again, we use the power rule and the sum/difference rule of derivatives:
x''(t) = 6t - 24
So, the acceleration of the particle is given by the function x''(t) = 6t - 24.
Therefore, the velocity function is x'(t) = 3t^2 - 24t + 21 and the acceleration function is x''(t) = 6t - 24.