Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Jan 26th 2009, 04:15 AM #1 Member   Join Date: Jan 2009 Posts: 80 Question on accelaration 2 (pls guide me,tq) A 40-kg girl and a 8.4-kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope the girl exerts a 5.2 N force on the sled pulling it toward her. a. What is the acceleration of the sled? b. What is the acceleration of the girl? c. How far from the girls initial position do they meet presuming the force remains constant and the no frictional forces act? Here is my working,am i right?Pls guide me.Thank you. a.the acceleration of the sled = a, F = m1 * a a = (F/m1) F = 5.2 N a = (5.2/40) = 0.13 m/s2 b.the acceleration of the girl = a1, F = m2 * a1 a1= (F/m2) = (5.2/8.4) = 0.62 m/s2 c.The distance covered by the sled when the girl pulls it is S = ut + (1/2)at2 S = 15 m,u = 0 m/s and a = 0.13 m/s2 15 = 0 * t + (1/2) * 0.13 * t2 t2= (15/0.065) = (15000/65) t = 15.2s The distance from the girl's initial position when they meet is S1 = ut + (1/2)a1t2 u = 0 m/s S1 = 0 * 15.2 + (1/2) * 0.62 * (15.2)2 S1 = 71.6 m Last edited by jonbrutal; Jan 26th 2009 at 04:45 AM.   Jan 26th 2009, 06:04 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 For the acceleration in part a) and b), it is better to have one being positive and one being negative as they are in opposite directions. For part c), you should note that both the girl and the sled are moving so it is not correct for you to find the time be at a distance of 15m. You should let the distance from the girl's position where they met be x m and on the other hand the distance from the sled's position where they met would be 15-x m, then you can use equation of motion and solve for x Last edited by werehk; Jan 26th 2009 at 06:07 PM.   Jan 26th 2009, 07:29 PM #3 Member   Join Date: Jan 2009 Posts: 80 Is this what you meant? I tried what u said,should i find the time first? This is what i did, The distance of the girl from initial position x = 1/2 (o.62)t*2=0.31t*2 the distance of the sled from the initial position 15 - x = 1/2(0.13)t*2=0.065t*2 SHould i equal both equations and find the value of x??Like this x = 0.31t*2 t*2= x/0.31 15-x=0.065t*2 t*2= (15-x)/0.065 (15-x)/0.065 = x/0.31 0.065x=4.65-0.31x x = 12.4 m Please help to explain,thank you for your help & support.I will make the accelaration of the sled negative. Last edited by jonbrutal; Jan 26th 2009 at 08:18 PM.  Tags accelaration, guide, pls, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post JJoll Advanced Optics 4 Nov 14th 2013 04:12 AM jonbrutal Kinematics and Dynamics 3 Jan 27th 2009 11:39 PM jonbrutal Kinematics and Dynamics 4 Jan 26th 2009 07:37 PM oceanmd Kinematics and Dynamics 2 Oct 20th 2008 08:58 AM SengNee Kinematics and Dynamics 3 May 27th 2008 04:52 AM