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:

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