Q. 90 - Calculus

Question

Ian is a bit worried about taking a fall into a crevasse while carrying a heavy pack and towing a heavy sled. He does some tests on an old rope, dropping from a tree in his backyard as shown below.

He cannot measure the force on the rope, but he takes a video, from which he can find his position (in feet below the tree limb) at time $t>1.2$ seconds as

$x\left(t\right) = 20+7e^{-0.25t}\sin (4.7t-5.8).$

(a) Weight is a force, given by mass times acceleration. Ian weighs 160 pounds, and the acceleration due to gravity that causes his weight is 32 feet per square second. What is Ian’s mass? (The units are called “slugs.”)

(b) Recalling that acceleration is the second derivative of position x(t), what is the force on Ian at any time $t>1.2$?

(c) Use a graphing calculator or other graphing utility to make a graph of Ian’s acceleration over time. When is the upward force on Ian the greatest? Note that since we are measuring distance below the tree limb, in this situation an upward force is negative. What is that force?

Step-by-Step Solution

Answer

1. Thus, the person I weighs 16.4 pounds or 7.4 kilograms.
2. 
   
   

Part (a) Step 1: Given Information.

At time $t>1.2$ seconds, the position on the rope is determined by the expression $x\left(t\right)=20+7e^{-0.25t}\sin (4.7t-5.8)$. Rope force cannot be measured. 160 pounds and 32 feet per square second of acceleration are attributed to the person.

Part (a) Step 2: Calculation

Let's use $M$ for the person's weight and $m$ for the person's mass.

$M=72.57$ kilogram width="45" style="max-width: none; vertical-align: -4px;" $=160$ pounds

$g=32$ per square metre in dollars metres per square metre are $9.7536$

Since mass times acceleration equals weight, weight is a force.

$$\begin{gathered} F=M=mg \\ m=\frac{M}{g} \\ m=\frac{72.57}{9.7536} \\ m=7.44 \end{gathered}$$

Therefore, the person i weighs 16.4 pounds or 7.4 kilograms.

Part (b) Step 1: Given information

At time $t>1.2$ seconds, the position on the rope is determined by the expression $x\left(t\right)=20+7e^{-0.25t}\sin (4.7t-5.8)$. Rope force cannot be measured. 160 pounds and 32 feet per square second of acceleration are attributed to the person.

Part (b) Step 2: Calculation

Take into account the acceleration, which is $x\left(t\right)$ second derivative.

$$\begin{gathered} x\left(t\right)=20+7e^{-0.25t}\sin (4.7t-5.8) \\ {x}^{\text{'}}\left(t\right)=7e^{-0.25t}[4.7\cos (4.7t-5.8\left)\right]+7(-0.25)\sin (4.7t-5.8)e^{-0.25t} \\ {x}^{\text{'}\text{'}}\left(t\right)=-7(4.7{)}^2{e}^{-0.25t}\sin (4.7t-5.8)-7(4.7\left)\right(0.25\left)\cos \right(4.7t-5.8)e^{-0.25t} \\ -7(0.25\left)\right(4.7\left)e^{-0.25t}\cos \right(4.7t-5.8)+7\left({}_{)}{}^{0.25}\sin (4.7t-5.8)e^{-0.25t}\right. \\ {x}^{\text{'}\text{'}}(t)=-154.63e^{-0.25t}\sin (4.7t-5.8)-8.225\cos (4.7t-5.8)e^{-0.25t} \\ -8.225e^{-0.25t}\cos (4.7t-5.8)+0.4375\sin (4.7t-5.8\left)e^{-0.25t} \\ {x}^{\text{'}\text{'}}\right(t)=e^{-0.25t}\sin (4.7t-5.8\left)\right[-154.63+0.4375] \\ +e^{-0.25t}\cos (4.7t-5.8\left)\right[-8.225-8.225\left] \\ {x}^{\text{'}\text{'}}\right(t)=-154.2e^{-0.25t}\sin (4.7t-5.8)-16.5e^{-0.25t}\cos (4.7t-5.8) \end{gathered}$$

Mass times acceleration thus yields the force acting on The Person I at any given time. i.e.

$$\begin{gathered} F\left(t\right)=m{x}^{\text{'}\text{'}}\left(t\right) \\ F\left(t\right)=7.4\left[-154.2e^{-0.25t}\sin (4.7t-5.8)-16.5e^{-0.25t}\cos (4.7t-5.8)\right] \end{gathered}$$

Part (c) Step 1: Given information

At time $t>1.2$ seconds, the position on the rope is determined by the expression $x(t)=20+7 e-0.25 t sin (4.7 t-5.8)$. Rope force cannot be measured. 160 pounds and 32 feet per square second of acceleration are attributed to the person.