Easy money (basic calc)

a) In this question, we have a given equation for the distance $s$ of an object given by:

$$s(t) = 7t^2 e^{-0.4t}$$

i) Find the acceleration when $t=3$. Acceleration is the second derivative of position, so to find the acceleration, we differentiate $s(t)$ twice and plug in $t=3$.

We'll do this in order, each derivative can be found by application of the chain and product rules. The first derivative $s'(t)$ is given by:

$$s'(t) = e^{-0.4 t} t (14.\, -2.8 t)$$

The second derivative, acceleration is given by:

$$s''(t) = e^{-0.4 t} \left(1.12 t^2-11.2 t+14.\right)$$

Plugging in $t=3$, we get:

$$\boxed{s''(3) = -2.867}$$

ii) The maximum value of the distance beyond $t=0$ can be found by finding the critical points of $s(t)$ to find a local maximum. These critical points are the points where the velocity, $s'(t)$ is equal to zero. We found the velocity above, so we need to solve for $s'(t) = 0$:

$$s'(t) = e^{-0.4 t} t (14.\, -2.8 t) = 0 \implies t = 0,5$$

Since we only care about the maximum in $t>0$, we have our solution as $t = 5$. To double check that this is a maximum, we can use the second derivative test: if $s''(5) < 0$, then it is a local maximum. Plugging in, we get $s''(5) = -1.89$, so it is indeed a local max. Therefore we plug $t=5$ into $s$ to get the value of the maximum position:

$$\boxed{s(5) = 23.68}$$ 

b) Now we are given an acceleration:

$$a = 8t + 7$$

With the following initial conditions on velocity and position: $v(0) = 4$ and $s(0) = 12$. To find equations for $v$ and $s$, we integrate the acceleration:

$$v(t) = \int 8t + 7\,dt = 4t^2 + 7t + C$$

We plug in $v(0) = 4$ to find $C = 4$. Therefore:

$$\boxed{v(t) = 4t^2 + 7t + 4}$$

Now we do the same process but for position:

$$s(t) = \int 4t^2 + 7t + 4\,dt = \frac{4}{3}t^3 + \frac{7}{2}t^2 + 4t + C$$

We use our initial condition $s(0) = 12$ to get $C = 12$ and arrive at our final answer:

$$\boxed{s(t) = \frac{4}{3}t^3 + \frac{7}{2}t^2 + 4t + 12}$$