Detailed Answer

a) The car changes direction when its velocity changes sign, in other words, where it has an x-intercept and changes sign. We can use the GDC to find this first intercept, which is at \( t = 6.03 \).

b) When the car's displacement decreases, it is moving closer to the origin. This means after the initial positive velocity (movement to the right), the car will move to the left when the velocity is negative. We can use the GDC to find the other x-intercept, which occurs at \( t = 8.83 \). Hence, the interval is \( 6.03 \leq t \leq 8.83 \).

c) To find displacement from velocity, we need to integrate the velocity function. Since the question asks for displacement, not distance, we just need to integrate the function as it is:

\[ \int_0^8 \left( 2\cos(0.5t) - 0.5t + 5 \right) dt = 21.0 \; \text{(with GDC)} \]

Detailed Answer

a) \( v(1) = e^{-2} + 5 = 5.14 \)

b) If we plot the graph in our GDC, it is clear to see that the maximum velocity will not occur where the derivative of the function is 0. After that point, the balloon's speed continues to increase. This means the maximum speed will occur at the end of the interval. Hence, \( v(4) = 20.0 \).

c) The acceleration is 0 when the derivative of the velocity function is 0, which occurs at \( t = 0.149 \). The distance covered up to this point is:

\[ \int_0^{0.149} \left( e^{-2t} + 5t^2 \right) dt = 0.134 \; \text{(with GDC)} \]

Detailed Answer

a) For this, we need to plot the graph provided in our GDC, and then find the x-coordinate of the first local maximum. This is where the derivative of the function will be 0, as the derivative of the velocity-time graph represents acceleration.

b) When the car is instantaneously at rest, its speed is 0, meaning the function has an x-intercept. The third x-intercept of the graph is at \( t_2 = \frac{9\pi}{4} \).

c) We need to find the distance covered between these two times. However, we can see that the car changes directions during this interval, as its velocity is both positive and negative. Since we are asked for the distance, we need to find the total area under the curve. Therefore, we integrate the absolute value of the function:

\[ \int_{1.89}^{\frac{9\pi}{4}} \left| 2e^{-\frac{t}{2}} \sin\left(t - \frac{\pi}{4}\right) \right| dt = 1.05 \]

Detailed Answer

a) The ball first comes to rest when its velocity is 0. Since velocity is the derivative of the displacement function, we need to find where the derivative of the displacement function is 0. This can be done by plotting \( s(t) \) and finding the first time its tangent is horizontal, which occurs at \( t_1 = 1.57 \).

b) To find the distance covered, we need to integrate the absolute value of the velocity function. We can do this by plotting the derivative of the \( s(t) \) function in the GDC, then plotting the absolute value of it, and evaluating the area underneath the graph up until \( t_1 \).

\[ \int_0^{1.89} |v(t)| dt = 4.62 \]

Detailed Answer

a) The particle is at rest when its velocity is 0, which occurs at the x-intercepts of the graph. In the domain provided, the x-intercepts are at \( t = 3.33, 5.27, 9.61 \).

b) The particle changes direction when its velocity changes sign. The first time this happens is at \( t = 3.33 \) seconds. Since acceleration is the derivative of the velocity function, we have:

\[ a(3.33) = v'(3.33) = -1.89 \;\; \text{[GDC]} \]

c) The total distance traveled can be expressed as the integral of the absolute value of the velocity function:

\[ \int_0^{10} |e^{\cos{(t)}} + 2\sin{t}| dt = 19.1 \;\; \text{[GDC]} \]

Detailed Answer

a) The particle changes direction when its speed changes sign, meaning the function has a zero. When the graph is plotted, it can be seen that there is an x-intercept at \( t = \pi \).

b) To solve this, we can plot the derivative of the function in our GDC to obtain the acceleration-time graph. Then, we plot the line \( y = 1.23 \) and find the intersection of these two functions, which occurs at \( t = 0.625 \) and \( t = 1.59 \).

c) The speed is the magnitude of the velocity, meaning it neglects direction. The speed is largest at the end of the interval, at \( t = \frac{7\pi}{6} \). Using graph trace, we find the value of the acceleration-time graph at this time to be \( -6.99 \, \text{ms}^{-2} \).

Detailed Answer

a) The particle is at rest when its velocity is 0, meaning it has an x-intercept. This function's zero is at \( t = 3.55 \, \text{s} \).

b) To find the displacement, we need to integrate the velocity function:

\[ \int_0^3 t^{2} \sin t + 5 \, dt = 20.8 \]

c) To find the acceleration, we need to evaluate the derivative of the velocity function at \( t = 1 \):

\[ a(1) = v'(1) = 2.22 \]

Detailed Answer

a) To solve this, we need to integrate velocity, as that will give displacement:

\[ \int_0^2 \frac{-4t^{3}+21t^{2}-18t+44}{6} \, dt = \left[-\frac{1}{6}t^4 + \frac{7}{6}t^3 - \frac{3}{2}t^2 + \frac{22}{3}t \right]_0^2 = \frac{46}{3} \]

b) To find the acceleration, we need to take the derivative of velocity:

\[ a(t) = v'(t) = -2t^2 + 7t - 3 \]

c) We have a maximum velocity when the acceleration of the object is 0 (the function of velocity has a maximum), which we can solve for:

\[ a(t) = -2t^2 + 7t - 3 = 0 \\ = 2t^2 - t - 6t + 3 = 0 \\ = (2t - 1)(t - 3) = 0 \]

From this, we get that \( t = 0.5 \) or \( t = 3 \). We can see from the function provided that at the later time, the velocity is larger, hence the largest speed is at \( t = 3 \). The speed at this time is:

\[ v(3) = \frac{-4(3)^3 + 21(3)^2 - 18(3) + 44}{6} = \frac{71}{6} \]

d) Since our velocity was always positive, we do not need to take the absolute value of the function, hence it is simply:

\[ \int_0^4 \frac{-4t^{3}+21t^{2}-18t+44}{6} \, dt \]