A point moving on a coordinate line has position function s. Find the velocity and acceleration at the t, and describe the motion of the point during the indicated time interval
s(t)=-2t^3+15t^2-24t-6 [0,5]
v = ds/dt = -6 t^2 + 30 t -24
a = dv/dt = -12 t + 30
at t = 0 the point is at -6
at t = 1 it is at -17
at t = 2 it is at -10
etc until
at t = 5 it is at -1
graph that
the velocity changed sign in there, look for where v = 0
-6 t^2 +30 t -24 = 0
-t^2 + 5 t - 4 = 0
t^2 -5 t + 4 = 0
(t -4)(t-1) = 0
at t = 1 and at t = 4 the direction reverses
at t = 0 the velocity is -24 so it starts out at -6 moving south fast
at t = 1 it reverses course and starts up from -17 etc
at t = 4 it is at 10 and starts down toward reaching 1 at t = -5
To find the velocity and acceleration at a given time t, you'll need to differentiate the position function s(t) with respect to time. Let's start by finding the velocity function by taking the derivative of s(t).
1. Take the derivative of s(t):
s'(t) = d/dt (-2t^3 + 15t^2 - 24t - 6)
To differentiate this function, we can apply the power rule of differentiation:
s'(t) = (-2 * 3t^2) + (15 * 2t) - 24
Simplifying further:
s'(t) = -6t^2 + 30t - 24
This gives us the velocity function v(t) = s'(t) = -6t^2 + 30t - 24.
Now, let's find the acceleration function by taking the derivative of the velocity function.
2. Take the derivative of v(t):
v'(t) = d/dt (-6t^2 + 30t - 24)
Again, applying the power rule of differentiation:
v'(t) = (-6 * 2t) + 30
Simplifying further:
v'(t) = -12t + 30
This is the acceleration function a(t) = v'(t) = -12t + 30.
Now we have the velocity function v(t) = -6t^2 + 30t - 24 and the acceleration function a(t) = -12t + 30.
To describe the motion of the point during the time interval [0, 5], we can evaluate the velocity and acceleration at both endpoints.
3. Evaluate the velocity at t = 0 and t = 5:
v(0) = -6(0)^2 + 30(0) - 24 = -24
v(5) = -6(5)^2 + 30(5) - 24 = 126
The velocity at t = 0 is -24 units/time, while at t = 5 is 126 units/time. This tells us that at the start, the point was moving backward with a speed of 24 units/time, and at t = 5, it is moving forward with a speed of 126 units/time.
4. Evaluate the acceleration at t = 0 and t = 5:
a(0) = -12(0) + 30 = 30
a(5) = -12(5) + 30 = -30
The acceleration at t = 0 is 30 units/time^2, while at t = 5 is -30 units/time^2. This indicates that at the beginning, the point is accelerating forward with a magnitude of 30 units/time^2, while at t = 5, it is decelerating or accelerating backward with a magnitude of 30 units/time^2.
Based on these findings, during the time interval [0, 5], the point's motion goes through the following phases:
- Initially, it moves backward, accelerating forward with a decreasing rate.
- Then, it stops momentarily at t = 0, changing its direction.
- After that, it starts moving forward, accelerating backward with an increasing rate.
- Finally, it stops again at t = 5, changing its direction again.
Please note that "units/time" can represent any units you have used to measure the coordinate line.