Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old 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.
GrahamJamesK is offline   Reply With Quote
Old Mar 14th 2011, 06:10 AM   #2
Physics Team
 
Unknown008's Avatar
 
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?
Unknown008 is offline   Reply With Quote
Old 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
GrahamJamesK is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
accelerations, constant acceleration, determining, kinematics, linear motion, particle, solved, stages



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
[SOLVED] Dymanics of a particle Quacky Kinematics and Dynamics 1 Sep 23rd 2010 03:25 PM
[SOLVED] Particle of mass m has momentum of magnitude mc olyviab Quantum Physics 1 Apr 1st 2010 05:54 PM
Describe the main stages of an active noise control system. George321 Waves and Sound 0 Feb 14th 2010 05:14 PM
Accelerations of Objects fireemblem13 Kinematics and Dynamics 1 Dec 29th 2009 09:57 PM
Determining the angular and linear accelerations of a Rigid Body Arrowstar Advanced Mechanics 0 Aug 8th 2009 05:42 PM


Facebook Twitter Google+ RSS Feed