My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By Country Boy
Reply
 
LinkBack Thread Tools Display Modes
November 16th, 2015, 10:14 AM   #1
Newbie
 
Joined: Aug 2015
From: USA

Posts: 10
Thanks: 0

Deriving Equations for Laws of Motion

Guys I am trying to teach myself calculus, and I have stumbled across these manuals my dad had and there are a few problem series that I need help with.

Any suggestions or direction to help outside the forum is appreciated.

1) Suppose a body "A" does not begin at rest, but instead starts with an initial velocity of 10 meters per second downward at the point B at time t=0. Again let s be the distance AB. Use calculus to show that

(a) if the acceleration is equal to a constant g=9.8 meters per second per second, then

s= 10t + 4.9t^2

(b) if s= 10t + 4.9 t^2, then the acceleration is equal to 9.8meters per second per second and the velocity

v=10 + 9.8t

2) Suppose that the body begins at a position 1 meter above the point B with an upward velocity of 15 meters per second. Find an equation for s in terms of t.

3) Suppose that the body begins at rest at the point B , but now suppose that the force is not constant. Instead, suppose that the body is acted on by a force causing a downward acceleration

a=2t + 1

Find an equation for s in terms of t

4) Suppose that acceleration is given by the equation

a= sin t

(a) If the body begins at point B at rest at ti t=0, find an equation for s in terms of t

(b) If the body begins at point B with a downward velocity of 5 meters per second, find an equation for s in terms of t.
Hey can you help me deriving these equations?

Suppose a body "A" does not begin at rest, but instead starts with an initial velocity of 10 meters per second downward at the point B at time t=0. Again let s be the distance AB. Use calculus to show that

(a) if the acceleration is equal to a constant g=9.8 meters per second per second, then

s= 10t + 4.9t^2

(b) if s= 10t + 4.9 t^2, then the acceleration is equal to 9.8meters per second per second and the velocity

v=10 + 9.8t

2) Suppose that the body begins at a position 1 meter above the point B with an upward velocity of 15 meters per second. Find an equation for s in terms of t.

3) Suppose that the body begins at rest at the point B , but now suppose that the force is not constant. Instead, suppose that the body is acted on by a force causing a downward acceleration

a=2t + 1

Find an equation for s in terms of t

4) Suppose that acceleration is given by the equation

a= sin t

(a) If the body begins at point B at rest at time t=0, find an equation for s in terms of t.

(b) If the body begins at point B with a downward velocity of 5 meters per second, find an equation for s in terms of t.

Last edited by skipjack; November 16th, 2015 at 06:29 PM.
Xaleb is offline  
 
November 16th, 2015, 10:33 AM   #2
Newbie
 
Joined: Aug 2015
From: USA

Posts: 10
Thanks: 0

I think I understand but I have such low confidence. This was what I did for the first one:

I think I am taking integral from 0 to v. I have no idea how to represent that on here:

v= gdt
∫vdt= ∫([10,v])g(t)dt
s= ∫g (v) dt - ∫g (10) dt
= g∫vdt -g∫10dt
= s - 10t
= 1/2gt^2 +10t (Am I allowed to make this substitution?)
= 1/2 (9.80) t^2 +10t
= 4.9t^2 + 10t


s = 10t +4.9t^2
d/ds = d/dt(10t + 4.9t^2)
v = 10d/dt(t) + 4.9d/dt(t^2)
= 10 + 4.9(2t)
= 10 + 9.8t

Are these correct? If so, can you help me with the others?

Last edited by Xaleb; November 16th, 2015 at 10:38 AM.
Xaleb is offline  
November 16th, 2015, 02:26 PM   #3
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Quote:
Originally Posted by Xaleb View Post
I think I understand but I have such low confidence. This was what I did for the first one:

I think I am taking integral from 0 to v. I have no idea how to represent that on here:

v= gdt
Since acceleration is the derivative of velocity with respect to time, dv/dt= a.
The acceleration due to gravity is -g (negative because you are measuring height upward and acceleration is downward) so dv/dt= - and so
dv= -g dt.

Quote:
∫vdt= ∫([10,v])g(t)dt
Almost. $\displaystyle \int dv= \int -g dt$. Also your limits of integration on the right make no sense. "10" and "v" are values of speed, not time. They cannot be limits of integration for integration with respect to time. They could be limits of integration for the integral on the left. You would do better to do this as an indefinite integral, v= -gt+ C, and use the initial condition, that v(0)= -10 (again, negative because the initial speed is downward) to get C= -10: v= -gt- 10

Quote:
s= ∫g (v) dt - ∫g (10) dt
Writing "g(v)" makes it look like g is a function of t - just leave it as gv
But v= ds/dt so ds= vdt= (-gt- 10)dt. Integrating both sides $\displaystyle v= -(g/2)t^2- 10t+ C$

Quote:
= g∫vdt -g∫10dt
= s - 10t
= 1/2gt^2 +10t (Am I allowed to make this substitution?)
No, because s is not equal to (1/2)gt^2! It is s you are trying to find.

Quote:
= 1/2 (9.80) t^2 +10t
= 4.9t^2 + 10t


s = 10t +4.9t^2
d/ds = d/dt(10t + 4.9t^2)
v = 10d/dt(t) + 4.9d/dt(t^2)
= 10 + 4.9(2t)
= 10 + 9.8t

Are these correct? If so, can you help me with the others?
It is not clear what "C" in $\displaystyle s= -(1/2)gt^2- 10t+ C$ should be because it is not clear what the initial height is. You say "Suppose a body "A" does not begin at rest, but instead starts with an initial velocity of 10 meters per second downward at the point B at time t=0." If you are taking point B to be s= 0 then C will be 0 and the distance from B to A will be $\displaystyle s= -(1/2)gt^2- 10t$. That is negative because A is below B.
Thanks from topsquark

Last edited by skipjack; November 16th, 2015 at 06:37 PM.
Country Boy is offline  
November 17th, 2015, 03:08 PM   #4
Newbie
 
Joined: Aug 2015
From: USA

Posts: 10
Thanks: 0

Thank you so much.
Xaleb is offline  
November 19th, 2015, 06:17 AM   #5
Newbie
 
Joined: Sep 2015
From: New york

Posts: 17
Thanks: 4

Order yourself a copy of "Calculus Made Easy" by Silvanus P Thompson. Quickest way to learn basic calculus.
davidmoore63 is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
deriving, equations, laws, motion



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
deriving equations harley05 Differential Equations 1 April 10th, 2014 03:44 AM
Equations of motion of the classical 3 body problem spacey Physics 16 April 16th, 2013 11:06 AM
Deriving Equations of a Periodic Function Tetra Algebra 3 December 11th, 2009 04:19 AM
Help deriving algebraic equations fzeropro Algebra 1 July 23rd, 2008 08:54 AM
Centrifuge Motion Equations Kerry Applied Math 0 July 23rd, 2008 05:52 AM





Copyright © 2018 My Math Forum. All rights reserved.