A particle is traveling along a one-dimensional path (such as a number line). The position of the particle is
governed by the time function x(t) ƒ 3t 4 ƒ{16t3 ƒy18t 2 ƒy 2 , where t is in minutes and 0 „T t „T 5 . Answer the
following questions.
a) At what times is the particle stationary?
b) For which time intervals is the particle moving in a positive direction? A negative direction?
c) What is the particle¡¦s most positive position? Most negative position?
d) What is the particle¡¦s displacement? What is the total distance the particle has traveled?
e) When does the particle¡¦s acceleration undergo a sign change? What is the particle¡¦s acceleration at the
times when the particle is stationary?
f) Sketch a graph (on a t-x coordinate plane) of the particle¡¦s position using the information above.
To answer the questions, we need to analyze the given function x(t) = 3t^4 - 16t^3 + 18t^2 - 2 over the interval 0 ≤ t ≤ 5.
a) The particle is stationary when its velocity is zero, which can be determined by finding the points where the derivative of the position function is zero. So, we need to find x'(t) = 12t^3 - 48t^2 + 36t. Setting x'(t) = 0, we can factor out 12t to get 12t(t^2 - 4t + 3) = 0. Solving for t, we find t = 0, t = 2, and t = 3. Therefore, the particle is stationary at times t = 0, t = 2, and t = 3.
b) The particle is moving in a positive direction when its velocity is positive, which corresponds to positive values of x'(t). We can determine the intervals by analyzing the sign of x'(t) over the interval 0 ≤ t ≤ 5.
To do this, we can analyze the sign of each term in x'(t) separately. When t < 0, all terms are negative, so x'(t) is negative. For 0 < t < 2, the first two terms are positive, and the last term is negative, so x'(t) is positive. For 2 < t < 3, the second term becomes negative, so x'(t) is negative. For t > 3, all terms are positive, so x'(t) is positive.
Therefore, the particle is moving in a positive direction for the interval 0 < t < 2 and for t > 3. It is moving in a negative direction for the interval 2 < t < 3.
c) To find the most positive and most negative positions, we need to find the maximum and minimum values of x(t) on the interval 0 ≤ t ≤ 5.
Taking the derivative of x(t) and setting it to zero, we find x'(t) = 12t^3 - 48t^2 + 36t = 0. Factoring out 12t, we get 12t(t^2 - 4t + 3) = 0. Solving for t, we find t = 0, t = 2, and t = 3.
Now, we need to evaluate x(t) at these critical points and the endpoints of the interval.
x(0) = 3(0^4) - 16(0^3) + 18(0^2) - 2 = -2
x(2) = 3(2^4) - 16(2^3) + 18(2^2) - 2 = 2
x(3) = 3(3^4) - 16(3^3) + 18(3^2) - 2 = 35
x(5) = 3(5^4) - 16(5^3) + 18(5^2) - 2 = 128
Therefore, the particle's most positive position is x = 128, and the most negative position is x = -2.
d) The particle's displacement is the difference in its position between the initial and final times. So, the displacement is x(5) - x(0) = 128 - (-2) = 130 units.
The total distance traveled by the particle is the sum of all the absolute values of the changes in position. So, the total distance is |128 - (-2)| + |2 - (-2)| + |35 - 2| = 128 + 4 + 33 = 165 units.
e) The particle's acceleration is the derivative of its velocity, which is the second derivative of the position function. So, we need to find x''(t) = 36t^2 - 96t + 36.
To determine when the acceleration undergoes a sign change, we need to find when x''(t) = 0 and analyze the sign changes around those points.
Solving 36t^2 - 96t + 36 = 0, we get t = 1 and t = 2. Analyzing the sign of x''(t) around these points, we find that x''(t) changes sign from positive to negative at t = 1, and from negative to positive at t = 2.
The particle's acceleration at the times when it is stationary (t = 0, t = 2, t = 3) can be found by evaluating x''(t) at those points:
x''(0) = 36(0^2) - 96(0) + 36 = 36
x''(2) = 36(2^2) - 96(2) + 36 = 0
x''(3) = 36(3^2) - 96(3) + 36 = 0
Therefore, the particle's acceleration is 36 when it is stationary at t = 0, and it is zero when it is stationary at t = 2 and t = 3.
f) To sketch a graph of the particle's position on a t-x coordinate plane, plot the points (t, x(t)) for each value of t in the interval 0 ≤ t ≤ 5. You can also plot additional points to determine the behavior of the curve. Connect the points with a smooth curve.
Note: Since I am a text-based AI, I cannot provide a graphical representation here. Please refer to the positional values calculated above and plot them on a graph to visualize the particle's position over time.