Physics Help Forum Phys Mechanics Kinematics

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 22nd 2014, 11:35 AM #1 Junior Member   Join Date: Sep 2014 Posts: 11 Phys Mechanics Kinematics A block is attached to the wall by some complicated spring-like device. It starts at the point x=A at time t=0 moving to the right with a velocity whose magnitude varies with time according to v(t)=c1t-c2t^2 where c1 and c2 are known constants. How far will it travel before starting back to left? Please help, I thought you would just take the integral of v(t)=c1t-c2t^2 and that gave me x(t)= c1t^2/2 -c2t^3/3 +A but the answer is x(t)=1/6 c1^3/c2^2 +A
 Sep 22nd 2014, 12:21 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 You are half way there. You have an equation for position as a function of time, but you haven't yet figured out what the value of t is. The object starts to move backward when v(t) = 0. Come up with that value, then substitue in for t in your equation and you shoud get the answer.
 Sep 22nd 2014, 12:58 PM #3 Junior Member   Join Date: Sep 2014 Posts: 11 Okay I am not sure I really understand. Do you mean 0=c1t-c2t^2 and solve for t then plug that number (c1/c2) into the x(t) equation for t?
Sep 22nd 2014, 01:04 PM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
 Originally Posted by NSB3 Okay I am not sure I really understand. Do you mean 0=c1t-c2t^2 and solve for t then plug that number (c1/c2) into the x(t) equation for t?
Precisely.

 Tags kinematics, mechanics, phys

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post nolanbradshaw01 Kinematics and Dynamics 1 Nov 20th 2015 04:38 AM amit6234623 Kinematics and Dynamics 1 Oct 8th 2015 06:13 AM charlielife Thermodynamics and Fluid Mechanics 3 Jun 13th 2014 05:37 AM Aladdin Kinematics and Dynamics 8 Apr 3rd 2011 11:08 AM akshay1995 Kinematics and Dynamics 1 Dec 24th 2010 03:34 AM