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November 16th, 2015, 09:14 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 05:29 PM. 
November 16th, 2015, 09:33 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 09:38 AM. 
November 16th, 2015, 01:26 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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:
Quote:
But v= ds/dt so ds= vdt= (gt 10)dt. Integrating both sides $\displaystyle v= (g/2)t^2 10t+ C$ Quote:
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Last edited by skipjack; November 16th, 2015 at 05:37 PM.  
November 17th, 2015, 02:08 PM  #4 
Newbie Joined: Aug 2015 From: USA Posts: 10 Thanks: 0 
Thank you so much.

November 19th, 2015, 05:17 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.


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