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Kinematics and jelly

Kinematics is part of the Calculus unit in IB Math courses. It is expected that you know how to find derivatives, integrals (through all its techniques) and also how to interpret them. Kinematics questions are usually found in Paper 2 questions, and that is why I decided to create one which would fit a Paper 1 (shake things up a bit). When answering the following question, be careful about what you’ll use as the origin (is it the surface of the fluid? Is it where the particle is released?) and also make sure you review a few concepts as well (for example, the difference between displacement and distance travelled). Now, on to the question.

Question 🤔

[Maximum marks: 20]

A particle is released from a certain height into a bowl of a viscous newtonian fluid. The particle penetrates the fluid and its velocity can be described by the function $v(t) = t \cdot e^{-t}$ where $v$ is in $m/s$ .

i) Find the maximum velocity reached by the projectile. [3 marks]

ii) Using the first or the second derivative test show that you found a maximum value. [2 marks]

iii) Find $v''(t)$ and interpret its meaning. [2 marks]

iv) Show that the particle's velocity of the object approaches zero as $t \to \infty$ . [2 marks]

i) Use the information from the previous items to graph $v(t)$ . [3 marks]

ii) Considering that $S(0) = 0$ and that the object starts penetrating the jelly after one second, find $S(t)$ . [4 marks]

iii) Hence or otherwise find the distance travelled in the first second [2 marks]

iv) How deep does the object penetrate the jelly? [2 marks]

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Solution 🤓

a) i) The maximum velocity can be found when its derivative is equal to zero (which is when the acceleration is equal to zero). Differentiating $v(t)$ with respect to $t$ using the product rule,

v' (t) = - e^{-t} + t \cdot e^{-t} = e^{-t} \cdot (t-1)

Now, making the derivative equal to zero:

e^{-t} \cdot (1-t) = 0 \quad \therefore \quad t = 1 s

(Given $e^{-t}$ is never equal to zero). For $t = 1$ :

v(1) = 1 \cdot e^{-1} = \frac{1}{e} m/s

a) ii) Now we have two possibilities for showing that we actually got a maximum, and not a minimum.

Using the first derivative test

Our critical point (and possibly maximum point) found was when $t = 1 s$ . Using a value smaller (such as 0.5) and bigger (such as 2) than 1 allows us to establish the sign of the derivative and see whether the function is increasing or decreasing before and after $t = 1 s$ .

$v'(0.5) =e^{-0.5} \cdot (1 - 0.5) > 0$

(the function is increasing)

$v' (2) = e^{-2} \cdot (1-2) < 0$

(the function is decreasing)

Given that the function increases and then decreases, we have a maximum value at $t = 1 s$

Using the second derivative test

As the same suggests, we first need to find the second derivative:

v'(t) = e^{-t} \cdot (1-t)

v''(t) = -e^{-t}(1-t) + e^{-t} \cdot (-1) = e^{-t} (t-2)

When plugging in $t = 1$ on the second derivative, we get

v''(1) = e^{-1} (1-2) < 0

which is a negative value (negative concavity), indicating that we actually got a maximum at $t = 1$ .

a) iii) We found $v''(t)$ in the previous item. It represents the instantaneous rate of change of the acceleration of the particle at any specific point.

a) iv) Evaluating the following limit and using L’Hopital’s rule:

\lim_{t \to \infty} v(t) = \lim_{t \to \infty} t \cdot e^{-t} = \lim_{t \to \infty} \frac{t}{e^t} \stackrel{L'H}{=} \lim_{t \to \infty} \frac{1}{e^t} = 0

b) i) Graph of $v(t)$ :

b) ii) Integrating $S(t)$ with respect to t,

S(t) = \int v(t) \ dt = \int t \cdot e^{-t} \ dt

The next step requires the use of integration by parts:

\left[ u = t \qquad du = dt \qquad dv = e^{-t} \ dt \qquad v = - e^{-t} \right]

S(t) = - t \cdot e^{-t} + \int e^{-t} \ dt = - t \cdot e^{-t} - e^{-t} + C = - e^{-t}(t+1) + C

It is stated that $S(0) = 0$ , therefore we can find the value of the constant C:

S(0) = - e^{0} \cdot (1) + C = 0 \quad \therefore \quad C = 1

It follows that

S(t) = - e^{-t} \cdot (t+1) + 1

b) iii) There are two possibilities to solve the following exercise: you either integrate $v(t)$ from 0 to 1 or you may evaluate $S(1)$ (the latter providing the simpler solution). Both techniques will be used below.

Integrating (integration by parts, as shown on the previous exercise):

\begin{aligned}\int_{0}^{1} v(t) \ dt &= \int_{0}^{1} t \cdot e^{-t} \ dt\\[0.5cm] &=\left[ -e^{-t} \cdot (t+1) \right] \Big|_{0}^{1} \\[0.5cm]&= - 2e^{-1} + 1 \\[0.5cm]&= \left( 1 - \frac{2}{e} \right) m\end{aligned}

Evaluating S(1):

S(1) = -e^{-1} \cdot (-2) + 1 = \left( 1 - \frac{2}{e} \right) m

b) iv) As t approaches infinity, the distance travelled approaches

\begin{aligned}\lim_{t \to \infty} S(t) &= \lim_{t \to \infty} \left[ - e^{-t} \cdot (-t-1) + 1 \right] \\[0.5cm] &= 1 + \lim_{t \to \infty} \left( \frac{t+1}{e^t}\right) \stackrel{L'H}{=} 1 +\lim_{t \to \infty} \left( \frac{1}{e^t} \right)\\[0.5cm] &= 1 \ m\end{aligned}

which means that the object will penetrate $1 \ m$ of jelly (as $t \to \infty)$ .