 Physics Help Forum [SOLVED] Determining accelerations of a particle over several stages
 User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Mar 14th 2011, 12:12 AM #1 Banned   Join Date: Feb 2011 Location: Galway, Ireland Posts: 17 [SOLVED] Determining accelerations of a particle over several stages This is a question from an early problem sheet in my Applied Math course. Although I am familiar with these problems and am reasonably competent with the equations of motion, this question is catching me out, I've tried all the formulae and still cannot seem to derive sensible answers. Your help is appreciated. A particle travels a distance of 300 m in a straight line at an average speed of 4 m/s, going from rest with constant acceleration a1 for 10 s, then moving with constant speed and then coming to rest with a constant retardation a2 for the last 20 s of the motion. Using the constant acceleration formalism for each stage of the motion, ﬁnd what must be the values of a1 and a2. Deduce, also, how far the particle travels while accelerating, while travelling at a constant velocity and while decelerating. Ok here is my attempt so far. Equations: v=u+at v^2=u^2+2as s=ut+1/2at^2 a=deltav/deltat Known values: t_total=s/v_av=300/4=75s t_1=10 t_2=45 t_3=20 s=300m v_av=4ms^-1 It can be reasoned that as it takes twice the time to decelerate from the constant velocity as it takes to reach it, the deceleration is half the acceleration; i.e. a1=a2/2 Stage 1: u_1=0ms^-1 v_1=?ms^-1 t_1=10s s_1=?m a1=?ms^-2 Stage 2: u_2=xms^-1 v_2=zms^-1 t_2=75-(10+20)=45s s_2=?m a=0ms^-2 Here u=v as velocity is consant Stage 3: u_3=?ms^-1 v_3=0ms^-1 a_2=?ms^-2 s_3=?m t_3=20s I've tried using the average velocity to work out the final velocity of stage one, but i end up with: v_f=2*v_av-v_o=2*4-0=8ms^-1, and this isn't right. Using the average velocity as my final velocity in stage one yields poor results too, what am i missing? Last edited by GrahamJamesK; Mar 14th 2011 at 12:16 AM.   Mar 14th 2011, 06:10 AM #2 Physics Team   Join Date: Jun 2010 Location: Mauritius Posts: 609 The fact that a good diagram will most of the time make the toughest problems instantly seem easier Draw a Velocity time graph. For the first 10 s, there is a constant slope of gradient a1. For the next part, there is a line with gradient 0, for some unknown time t, and between t and (20-t) seconds later, there is a slope with a negative gradient, a2. What you have is a trapezium. The properties of a VT graph is that the gradient gives the acceleration, and the area under the graph gives the displacement. You got t = 45 s, meaning that on your graph, on the horizontal axis, you'll have (45+10) = 55 s elapsed since the start of the motion and t = 75 s for the whole trip. Using the area, you can get a1 and a2 quickly. The total area is 300 m, that is: Area = 0.5 * v * (75 + 45) You can get v. From there, work out a1 and a2 easily And then, everything becomes easy. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?   Mar 14th 2011, 10:03 AM #3 Banned   Join Date: Feb 2011 Location: Galway, Ireland Posts: 17 Thanks Jerry! I was no where near the idea of area, i'll keep this is mind the next time. Stage 1: u_1=0ms^-1 v_1=Area/0.2*(75+45)=300/60=5ms^-1 t_1=10s s_1=(0.5*10^2)/2=25m a1=v-u/t=5/10=0.5ms^-2 Stage 2: u_2=5ms^-1 v_2=5ms^-1 t_2=75-(10+20)=45s s_2=v*t=5*45=225m a_0=0ms^-2 Here u=v as velocity is consant Stage 3: u_3=5ms^-1 v_3=0ms^-1 a_2=(v-u)/t=(0-5)/20=-0.25ms^-2 s_3=5*20+(-0.25*20^2)/2=50m t_3=20s s1+s2+s3=25+225+50=300m=total distance travelled. Thank you for your help!! Graham  Tags accelerations, constant acceleration, determining, kinematics, linear motion, particle, solved, stages 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 Quacky Kinematics and Dynamics 1 Sep 23rd 2010 03:25 PM olyviab Quantum Physics 1 Apr 1st 2010 05:54 PM George321 Waves and Sound 0 Feb 14th 2010 05:14 PM fireemblem13 Kinematics and Dynamics 1 Dec 29th 2009 09:57 PM Arrowstar Advanced Mechanics 0 Aug 8th 2009 05:42 PM