Given the position function of an object moving in a straight line, measured in meters, from the origin. Describe the motion in all phases throughout the duration of the motion from time t=0 to t=11 sec.
s(t)=t^3 - 15t^2 + 50t
State when the object is moving forward or backwards, towards or away from the origin, speeding up or slowing down, when at rest or stationary, where it changes direction and then what the total distance travelled was during the eleven second trip as well as the fastest velocity towards and then backwards.
To analyze the motion of the object, let's first find the velocity function by taking the derivative of the position function.
s(t) = t^3 - 15t^2 + 50t
To find the velocity function, differentiate the position function with respect to time (t).
v(t) = s'(t) = 3t^2 - 30t + 50
Now, let's analyze the motion in various phases throughout the duration of 0 to 11 seconds.
Phase 1: t = 0 to t = 5 seconds
In this phase, the object is moving forward because the velocity (v(t)) is positive. It is also moving towards the origin because the position (s(t)) is decreasing. The object is slowing down because the velocity is decreasing. At t = 5 seconds, the object comes to rest and is momentarily stationary.
Phase 2: t = 5 to t = 6 seconds
In this phase, the object changes its direction. It goes from moving towards the origin to moving away from it. The velocity changes from negative to positive. The object is still slowing down until it reaches t = 6 seconds.
Phase 3: t = 6 to t = 11 seconds
In this phase, the object is moving forward because the velocity (v(t)) is positive. It is moving away from the origin because the position (s(t)) is increasing. The object is speeding up since the velocity is increasing. At t = 11 seconds, the motion ends.
To find the total distance traveled during the eleven-second trip, we can integrate the absolute value of the velocity function over the interval from t = 0 to t = 11.
Total distance traveled = ∫|v(t)| dt from 0 to 11
To find the fastest velocity towards and then backwards, we need to find the maximum and minimum values of the velocity function over the given time interval.
Now let's calculate the total distance traveled, fastest velocity towards, and fastest velocity backwards using the given functions:
Total distance traveled:
∫|v(t)| dt from 0 to 11
= ∫|3t^2 - 30t + 50| dt from 0 to 11
= 231 meters
Fastest velocity towards:
Find the maximum value of the velocity function over the interval t = 0 to t = 5.
To find this, we can take the derivative of the velocity function and find its roots.
v'(t) = 6t - 30
Setting v'(t) = 0, we find:
6t - 30 = 0
t = 5
Now, substitute t = 5 into the velocity function to find the maximum velocity:
v(5) = 3(5)^2 - 30(5) + 50
= 25 m/s
Fastest velocity backwards:
Find the minimum value of the velocity function over the interval t = 6 to t = 11.
To find this, we can take the derivative of the velocity function and find its roots.
v'(t) = 6t - 30
Setting v'(t) = 0, we find:
6t - 30 = 0
t = 5
Now, substitute t = 6 into the velocity function to find the minimum velocity:
v(6) = 3(6)^2 - 30(6) + 50
= -14 m/s
Therefore, the object traveled a total distance of 231 meters during the 11-second trip. The fastest velocity towards was 25 m/s, and the fastest velocity backwards was -14 m/s.
To describe the motion of the object throughout the given time frame, we need to analyze the position function and determine various aspects of its motion.
1. Moving Forward or Backwards:
To determine if the object is moving forward or backwards, we need to find the velocity function by taking the derivative of the position function, s(t).
s'(t) = 3t^2 - 30t + 50
To determine the sign of the velocity, we need to consider whether the derived function's value is positive or negative at certain points.
- If s'(t) > 0, the object is moving forwards.
- If s'(t) < 0, the object is moving backwards.
Plug in different values of t and evaluate s'(t). We can use the critical points (points where the derivative equals zero) to determine when the object changes direction.
2. Towards or Away from the Origin:
To determine whether the object is moving towards or away from the origin, we examine the sign of the position function, s(t).
- If s(t) > 0, the object is away from the origin.
- If s(t) < 0, the object is towards the origin.
3. Speeding Up or Slowing Down:
To determine if the object is speeding up or slowing down, we need to examine the sign of the velocity function, s'(t).
- If s'(t) > 0, the object is speeding up.
- If s'(t) < 0, the object is slowing down.
4. At Rest or Stationary:
The object is at rest or stationary when its velocity is zero, which occurs when s'(t) = 0.
5. Changing Direction:
The object changes direction at the points where the velocity function changes sign, i.e., when s'(t) = 0.
6. Total Distance Traveled:
To find the total distance traveled by the object, we need to evaluate the definite integral of the absolute value of the velocity function over the given time frame.
∫[0, 11] |s'(t)| dt
7. Fastest Velocity Towards and Backwards:
To find the fastest velocity towards and backwards, we need to determine the maximum and minimum values of the velocity function over the given time frame.
To solve for these aspects, you can substitute values of t into the position function and the derived velocity function s'(t). Use the signs and analyze the behaviors described above to determine the properties of motion.