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:

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

Detailed Answer

a) At \( t=0 \) seconds, the person has not thrown the stone yet, therefore it is still at the height of the mountain. We need to plug 0 into our given function. \( s(0) = -10(0)^2 + 15(0) + 10 = 10 \, \text{m} \).

b) Since the given displacement function is told to be above the sea level, we know that the sea level is at 0m, so we need to find the times when the stone is above 0m, more precisely above \( s(t) = 0 \). This we can do by finding the x intercepts of our function, which can be done either with the quadratic formula, or by factoring.

\[ s(t) = -10t^2 + 15t + 10 = -5(2t^2 - 3t +10) \]

\[ s(t) = -5(2t^2 +t -4t +10) = -5(t(2t+1)-2(2t+1)) \]

\[ s(t) = -5(2t+1)(t-2) \]

So from this we see we have roots at \( t = -\frac{1}{2} \) and \( t=2 \). Since we start at 0 seconds, the negative time doesn't make sense, hence we are above sea level from 0s to 2s.

c) We can take the derivative of our function to see where it takes its maximum value: \( s'(t) = -20t + 15 \), where this is 0, we will have the maximum height.

\[ -20t + 15 = 0 \implies t=0.75 \, \text{s} \]

At this time our height: \( s(0.75) = -10(0.75)^2 + 15(0.75) + 10 = 15.625 \, \text{m} \). We know this is our maximum as a quadratic can only have 1 max/min point, and this one is concave down, as our \( t^2 \) coefficient is negative, hence our found value must be a maximum.

Detailed Answer

a) \( s(0) = -20 \, \text{m} \), since it's negative, it's to the left.

b) Average velocity can be calculated by taking the total displacement and dividing it by the total time.

\[ v = \frac{s(10) - s(0)}{10 - 0} = \frac{(10(10)^2 - 20) - (-20)}{10} = 100 \]

c) To find the velocity and acceleration we need to take the first and second derivative.

\[ v(2) = s'(2) = 20(2) = 40 \, \text{m/s} \]

\[ a(2) = s''(2) = 20 \, \text{m/s}^2 \]

d) The particle is at the origin when its displacement is 0. Hence, \( s(t) = 0 = 10t^2 - 20 \), from this \( t = \pm \sqrt{2} \, \text{s} \), however, negative time in this context doesn't make sense.

Detailed Answer

To find the distance covered, we need to find the area underneath the velocity-time graph. We can do that by splitting it into triangles and rectangles. The areas summed up from left to right are calculated as follows:

\[ \text{Distance} = 9 + 12 + 2 + 16 = 39 \, \text{m} \]

Detailed Answer

a) To find the velocity function from the acceleration function, we must integrate:

\[ v(t) = \int 5 + e^{(5t +1)} \, dt = 5t + \frac{1}{5}e^{(5t +1)} + C \]

We can find \( C \) from the boundary condition, as we were told at the start it was at rest, so \( v(0) = 0 \), hence:

\[ C = - \frac{1}{5}e \]

b) To find the displacement function, we can take a similar approach, only now we need to integrate the velocity function:

\[ s(t) = \int \left(5t + \frac{1}{5}e^{(5t +1)} - \frac{1}{5}e \right) \, dt = \frac{5}{2}t^2 + \frac{1}{25}e^{(5t +1)} - \frac{1}{5}et + C \]

We can again use the boundary condition that \( s(0) = 0 \), hence:

\[ C = - \frac{1}{25}e \]

Detailed Answer

a)

b) To go from the velocity function to the acceleration function, we must take the derivative.

\[ a(t) = v'(t) = 10e^{-3t} - 30e^{-3t} = (10 - 30t)e^{-3t} \] (Using the product rule)

c) Velocity increases when our acceleration is positive. Graphing the acceleration function, it is clear that the function is positive when \( 0 \le t \le \frac{1}{3} \) s.

d) Given our graph from part (a), we can see the speed starts decreasing after the function reaches its maximum, so when \( t > \frac{1}{3} \) s.

Detailed Answer

a) We can find the velocity vector by taking the derivative of the x and y components of the parametric form one by one.

\[ \mathbf{v} = \begin{bmatrix} 5 \\ 2t + 5 \end{bmatrix} \]

b) The speed by definition is the magnitude of the velocity vector, hence:

\[ \text{speed} = \sqrt{5^2 + (2t+5)^2} = \sqrt{4t^2 + 20t + 50} \] (Simply finding the magnitude of the vector)

Detailed Answer

a) \( v(0) = \frac{250}{(0+2)^2} = 62.5 \, \text{m/s} \)

b) \( v(5) = \frac{250}{(5+2)^2} = 5.1 \, \text{m/s} \)

c) \( \int_{0}^{3} \frac{250}{(t+2)^2} \, dt = -250 \cdot \left[ \frac{1}{t+2} \right]^3_0 = 75 \, \text{m} \)

d) If we travel 25 more meters, it means in total the car travelled 100m, so we need to find the time interval when the above integral yields 100 meters:

\[ \int_{0}^{x} \frac{250}{(t+2)^2} \, dt = 100 \]

\[ -250 \cdot \left[ \frac{1}{t+2} \right]^x_0 = 100 \]

\[ -250 \cdot \left( \frac{1}{x+2} - \frac{1}{2} \right) = 100 \]

\[ -\frac{250}{x+2} = -25 \]

\[ 25x = 200 \]

\[ x = 8 \, \text{seconds} \]

Detailed Answer

a)

\[ \textbf{a} = \frac{d\textbf{v}}{d\textbf{t}} = 5\textbf{i} + 4\textbf{j} \]

\[ | \textbf{a} | = \sqrt{5^2 + 4^2} = \sqrt{41} \, \text{m/s}^2 \]

b)

\[ \textbf{r} = \int \textbf{v} \, dt = \int (5t+9)\textbf{i} + (4t-4)\textbf{j} \, dt \]

\[ = \left(\frac{5}{2}t^2 + 9t + C\right)\textbf{i} + \left(2t^2 - 4t + D\right)\textbf{j} \]

We were given, when \( t = 3 \), \( \textbf{r} = 5\textbf{i} \):

\[ \textbf{r}(3) = \left(\frac{45}{2} + 27 + C\right)\textbf{i} + \left(18 - 12 + D\right)\textbf{j} = 5\textbf{i} + 0\textbf{j} \]

From this we see \( C = -\frac{89}{2} \) and \( D = -6 \):

\[ \textbf{r}(t) = \left(\frac{5}{2}t^2 + 9t - \frac{89}{2}\right)\textbf{i} + \left(2t^2 - 4t - 6\right)\textbf{j} \]

\[ \textbf{r}(0) = \left(\frac{5}{2}(0)^2 + 9(0) - \frac{89}{2}\right)\textbf{i} + \left(2(0)^2 - 4(0) - 6\right)\textbf{j} \]

\[ = -\frac{89}{2}\textbf{i} - 6\textbf{j} \]

The distance will be the magnitude of this displacement vector:

\[ |\textbf{r}(0)| = \sqrt{\left(\frac{89}{2}\right)^2 + 6^2} = \frac{\sqrt{8065}}{2} \, \text{m} \]