Refer a friend and earn $50 when they subscribe to an annual planRefer Now

Get the answer to your homework problem.

Try Numerade Free for 30 Days

Like

Report

An object of mass 5 kg is released from rest 1000 m abovethe ground and allowed to fall under the influence of grav-ity. Assuming the force due to air resistance is proportionalto the velocity of the object with proportionality constant$$b=50 \mathrm{N}-\mathrm{sec} / \mathrm{m}$$, determine the equation of motion of theobject. When will the object strike the ground?

$t \approx 1019.46 \mathrm{sec}$

Calculus 2 / BC

Chapter 3

Mathematical Models and Numerical Methods Involving First-Order Equations

Section 4

Newtonian Mechanics

Differential Equations

Cath J.

October 19, 2020

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

02:43

A $$400-1 \mathrm{b}$$ obj…

02:18

An object of mass 8 kg is …

03:19

An object falling under th…

04:18

Physics An object is prope…

03:58

Free Fall. In Section 2.1,…

03:43

An object is propelled ve…

09:51

An object of mass $m$ is d…

02:01

An object of mass m is dro…

04:05

A projectile with mass of …

01:10

If the object in Problem 2…

So from the given information, we have been given the mass off five. So that's let's say that's M, which is the equal to five and B, which is given as 50. So from this we have the two forces, if one and if to in such a way that if one the first force will be m Times G, which in this case will simplify to five g andi we also have if two which is negative b times the velocity which is a function off t and this will lean simplify to negative 50 times the velocity as a function off t. But we also know that mm times the derivative off the two T is then the force in terms off T and V, and this can become our first equation good. Furthermore, we have that the total force will be if one plus if to which we can substitute with M G plus Negative B VT, which simplifies to five G minus 50 the tea from all the information that was given. And let's call this thing equation too. From here, we substitute equation too into Equation one, and we end up with in times the derivative off the velocity to t equal to five g minus 50 V, which is still a function off t. So if we substitute everything indeed it will be five times the derivative off the velocity to time equal to five G minus 50 V. And from here we could divide by five just to simplify but further and we end up with Devi. The derivative off velocity to time is equal to G minus teen the Now we go ahead and we separate the variables. So we put all the V's on the one side and the T on the other side, so that will end up to be one over G minus 10 V D. V and we times DT on the other side on the right hand side. And that's all we have in terms off T. And from here we integrate both sides. Now we integrate both sides so we have one integral off one over G minus teen v D. V equal to the derivative off the team. So on the left hand side, the integral off the fraction will be negative. 1/10 Lynn off G minus 10 v and on the right hand side that will be t plus the constant C one. I'm going thio times with the negative tin right through the equation. And I end up with Len off G minus 10 v equal to negative 10 t And if I multiply the C one with negative 10 it just gives me another constant so I can change the c one to a new constant C two From the definition off. Len, this thing can be rewritten as g minus t v equal to e the constant e to the power off. Negative 10 t plus c two. And from this we have G minus 10 V equal to and the E to the power off C two becomes another constant, which I can substitute with C three and we still have to eat to the power off. Negative 10 t Now I'm going to take the G to the other side and I have 10 negative 10 v equal to C three e to the power of negative T minus g. So this gives me the basic function off the by dividing again with the negative 10. So in other words, the velocity in terms off time will simplify, then to G divided by 10 plus C three e to the power off negative teen T or rather just see because we are. That is our final constant that we are going to work with. Since the object WAAS released from raced, we have the initial condition off V zero with the time is zero and that is also zero. So 40 equal to zero hour V is also equal to zero. And let's substitute this thing into the equation that we've obtained here in order to find the value off our constant C. So from that we will have that VT can be substituted by zero. We have G over 10 stole minus or plus C times E to the power off zero that simplifies to zero equal to G over 10 plus c. So therefore, C is equal to negative g over 10. So let's put all off this thing back into our VT equation. So we just substitute to see what we got here. So V t is equal to G over 10 Plotnick minus G over teen E to the power of negative tin T and just full simpler. First, we can take out the G over teen as a common factor and we have one minus eater above negative Tain t left just to simplify a bit. Okay, on we will call vesting equation three by integrating the velocity we confined in an equation for X t, which is the distance in terms off the time that was given. So if we integrate this that it will be the integral off VT two DT, which is the integral off G over 10 which is a constant one minus e two part negative team T in terms off t so we can take out the nine the G over teen and we only need to find the integral, then off one minus e to part of negative teen T D. T, which ends up to be G over 10 t plus one over Teen E to the power of negative teen T and remember to add the constant C. So if you look at the situation where it was dropped 1000 m, then we can take our initial distance as being zero and then our distance after time t will then be 1000 so we can take this initial value and substituted into this equation to find the value off the sea. So let's do that. That's been going to give us zero equal to G. Over. Teen T is equal to zero plus one over Teen E to the power of zero plus C. Let's simplify that will be zero equal to G over 100 plus c so they foresee the constant is negative g over 100. Substituting this back into the original equation off X t. We will really integrated. So we're going to just simplify this substitute see in the and we end up with X T eggs, which is a function off. T will then be equal to G over 10 times T plus 1/10 e to the power off. Negative 10 t minus G over 100 and just to clean it up a bit, this can be simplified to G over tain t plus G over 100 times e to the power off. Negative 20 minus one. Then finally, if we consider this last bit off the equation, we off the situation where the distance is 1000 m after time T we can use this information to find t so we substitute X t with 1000 thing and we calculate team. So that is 1000 equal to G over tain t plus G over 100 times e to the power of negative 10 T minus one and using the constant G as 981 which is the acceleration due to gravity. We can substitute that as well now, and we'll end up with 1000 equal to nine comma 8 1/10 t 10 and then Times T plus 9 +81 over 100 times e to the power off. Negative 10 T minus one. And from this we can simplify to 1000 equal to zero comma 98 won t plus zero comma 09 81 So I'm just simplifying the fractions into decimals. So that's negative. Teen T minus one stole. And from this we get 1000 is equal to zero common 98 won t minus zero comma 09 81 Because when I multiplied the zero comma 0981 with E to the power of negative team, T t is always positive. So it is small enough to neglect. And from here we have seen that 1000 plus zero comma +0981 is equal to zero common +98 won t and that simplifies that T is then approximately one 1019 comma 46 seconds.

View More Answers From This Book

Find Another Textbook

In mathematics, integration is one of the two main operations in calculus, w…

In grammar, determiners are a class of words that are used in front of nouns…

A $$400-1 \mathrm{b}$$ object is released from rest 500 ft above the ground …

An object of mass 8 kg is given an upward initial veloc-ity of 20 m/sec …

An object falling under the pull of gravity is acted upon by a frictional fo…

Physics An object is propelled vertically upward with an initial velocity of…

Free Fall. In Section 2.1, we discussed a model for an object falling toward…

An object is propelled vertically upward with an initial velocity of 20 met…

An object of mass $m$ is dropped at $t=0$ from the roof of a building of he…

An object of mass m is dropped from the roof of a building of height h. Whil…

A projectile with mass of $500 \mathrm{kg}$ is launched straight up from the…

If the object in Problem 2 is released from rest $$30 \mathrm{ft}$$ above th…

02:27

In Problems $9-20$ , determine whether the equation is exact. If it is, then…

05:27

In Problems $7-16,$ solve the equation.$$ \frac{d y}{d x}=3 x^{2}\left(1…

01:13

If the resistance in the RL circuit of Figure 3.13(a) is zero,show that the …

02:40

A brine solution of salt flows at a constant rate of 6 L/min into a large ta…

01:27

Show that $$C_{1} \cos \omega t+C_{2} \sin \omega t$$ can be written in the …

10:06

Use the method described in Problem 32 to show thatthe orthogonal trajec…

07:55

In Problems 13 and $14,$ find an integrating factor of the form $x^{n} y^{m}…

17:08

A nitric acid solution flows at a constant rate of 6 L/min into a large tank…

02:24

In Problems $17-26,$ solve the initial value problem.$$ \sqrt{y} d x+(1+…

02:37

In Problems $17-22,$ solve the initial value problem.$$ \frac{d y}{d x}+…

92% of Numerade students report better grades.

Try Numerade Free for 30 Days. You can cancel at any time.

Annual

0.00/mo 0.00/mo

Billed annually at 0.00/yr after free trial

Monthly

0.00/mo

Billed monthly at 0.00/mo after free trial

Earn better grades with our study tools:

Textbooks

Video lessons matched directly to the problems in your textbooks.

Ask a Question

Can't find a question? Ask our 30,000+ educators for help.

Courses

Watch full-length courses, covering key principles and concepts.

AI Tutor

Receive weekly guidance from the world’s first A.I. Tutor, Ace.

30 day free trial, then pay 0.00/month

30 day free trial, then pay 0.00/year

You can cancel anytime

OR PAY WITH

Your subscription has started!

The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

Receive weekly guidance from the world's first A.I. Tutor, Ace.

Mount Everest weighs an estimated 357 trillion pounds

Snapshot a problem with the Numerade app, and we'll give you the video solution.

A cheetah can run up to 76 miles per hour, and can go from 0 to 60 miles per hour in less than three seconds.

Back in a jiffy? You'd better be fast! A "jiffy" is an actual length of time, equal to about 1/100th of a second.