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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
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Quote:
 Originally Posted by Xaleb 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.

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. Tags deriving, equations, laws, motion Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post harley05 Differential Equations 1 April 10th, 2014 02:44 AM spacey Physics 16 April 16th, 2013 10:06 AM Tetra Algebra 3 December 11th, 2009 03:19 AM fzeropro Algebra 1 July 23rd, 2008 07:54 AM Kerry Applied Math 0 July 23rd, 2008 04:52 AM

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