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 September 1st, 2012, 08:07 PM #1 Senior Member     Joined: Jan 2012 Posts: 639 Thanks: 6 A particle moving in a straight line A particle moving in a straight line with uinform deceleration has a velocity of 40ms-1 at a point P, 20ms-1 at a point Q and comes to rest a point R where QR=50, calculate the (i) distance PQ; (ii) time taken to cover PQ (iii) time taken to cover PR. How do about calculating this?
 September 1st, 2012, 08:51 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs Re: A particle moving in a sraight line We are given: $\frac{dv}{dt}=-k$ where $0. Let time $t=0$ coincide with the particle being at P, which we will take to be the origin of our coordinate axis x. So, integrating, we have: $v(t)=-kt+c_1$ Integrating again, we find: $x(t)=-\frac{k}{2}t^2+c_1t+c_2$ We are also given: $x(0)=P$ $v$$t(P)$$=v(0)=40\text{ \frac{m}{s}}$ $v$$t(Q)$$=20\text{ \frac{m}{s}}$ $v$$t(R)$$=0\text{ \frac{m}{s}}$ So, we may use this to state: $v(0)=c_1=40$ $v$$t(Q)$$=-kt(Q)+40=20\:\therefore\:k=\frac{20}{t(Q)}$ $v$$t(R)$$=-kt(R)+40=0\:\therefore\:k=\frac{40}{t(R)}$ hence: $\frac{20}{t(Q)}=\frac{40}{t(R)}$ $t(R)=2t(Q)$ $x$$t(P)$$=x(0)=c_2=0$ $x(t)=-\frac{10}{t(Q)}t^2+40t$ $x$$t(R)$$-x$$t(Q)$$=50$ $x$$2t(Q)$$-x$$t(Q)$$=50$ $$$-\frac{10}{t(Q)}\(2t(Q)$$^2+40$$2t(Q)$$\)-$$-\frac{10}{t(Q)}\(t(Q)$$^2+40$$t(Q)$$\)=50$ $-40t(Q)+80t(Q)+10t(Q)-40t(Q)=50$ $t(Q)=5$ $t(R)=2t(Q)=10$ $x$$t(Q)$$=x(5)=-2(5)^2+40(5)=150$ Thus, we have found: (i) distance PQ = 150 m. (ii) time taken to cover PQ = 5 s. (iii) time taken to cover PR = 10 s.
 September 1st, 2012, 11:48 PM #3 Senior Member     Joined: Jan 2012 Posts: 639 Thanks: 6 Re: A particle moving in a straight line But apart from the way you did it, can we use the equations of motion in solving it?
 September 1st, 2012, 11:53 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs Re: A particle moving in a straight line I simply used the calculus (along with the definitions of acceleration as the time rate of change of velocity, and velocity as the time rate of change of displacement) to derive the relevant equations of motion needed to solve the problem. Calculus is not necessary to derive these, but it is computationally simpler. If you are already given equations of motion, then that is simpler still.
 September 2nd, 2012, 12:09 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs Re: A particle moving in a straight line For example, if we have access to the equations of motion governing uniform acceleration in a straight line, then: we may find the acceleration from: $a=\frac{v_f^2-v_i^2}{2x}=-\frac{$$20\text{ \frac{m}{s}}$$^2}{2$$50\text{ m}$$}=-4\text{ \frac{m}{s^2}}$ (i) distance PQ: $x=\frac{v_f^2-v_i^2}{2a}=\frac{$$20^2-40^2$$}{2(-4)}\:\text{m}=150\text{ m}$ (ii) time take to cover PQ: $t=\frac{v_f-v_i}{a}=\frac{20-40}{-4}\:\text{ s}=5\text{ s}$ (iii) time take to cover PR: $t=\frac{v_f-v_i}{a}=\frac{0-40}{-4}\:\text{ s}=10\text{ s}$
 September 2nd, 2012, 12:51 AM #6 Senior Member     Joined: Jan 2012 Posts: 639 Thanks: 6 Re: A particle moving in a straight line In this problem, which is final and initial velocity and why? I want to know, let me see how I can understand this concept better.
 September 2nd, 2012, 12:55 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs Re: A particle moving in a straight line In the movement from P to Q, 40 m/s is the initial velocity and 20 m/s is the final velocity. In the movement from Q to R, 20 m/s is the initial velocity and 0 m/s is the final velocity. In the movement from P to R, 40 m/s is the initial velocity and 0 m/s is the final velocity. You take the velocity at the initial position and time and that is the initial velocity, likewise the final velocity is the velocity at the final position and time.
 September 2nd, 2012, 04:21 AM #8 Senior Member     Joined: Jan 2012 Posts: 639 Thanks: 6 Re: A particle moving in a straight line Ok now that the (i) distance PQ is required: the initial velocity u at PQ =20ms-1; the final velocity v =0ms-1. Using v^2=u^2+2as 0=20^2+2*50a -400=100a a=-400/100=-4m/s can I veiw that this way; that as the particle is moving in a straight line with a deceleration of 40m/s, by the time it get to Q, it velocity has decrease further to 4m/s? If the initial velocity and final velocity at PQ are 20m/s and 0m/s respectively, why don't we use it to calculate the distance PQ?
September 5th, 2012, 07:44 AM   #9
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Re: A particle moving in a straight line

Quote:
 Originally Posted by Chikis A particle moving in a straight line with uniform deceleration has a velocity of 40ms-1 at a point P, 20ms-1 at a point Q and comes to rest a point R where QR=50, calculate the (i) distance PQ; (ii) time taken to cover PQ (iii) time taken to cover PR. How do about calculating this?
Here is another way of looking at it:

We could plot a velocity-time graph to represent the motion of the particle. We know that the gradient of a line on a velocity-time graph represents the acceleration/deceleration of the particle and since we're told that the particle decelerates uniformly from point P (with velocity of 40ms-1) to point Q (with velocity of 20ms-1) then comes to rest at point R. This means we have a straight line that slopes down.
[attachment=2c5cclog]MAAS Part1.JPG[/attachmentc5cclog]

Given the distance between QR is 50m, but we know we can find the distance between QR by determining the area under velocity-time line, in this case, we have a right triangle here.
[attachment=1c5cclog]MAAS Part 2.JPG[/attachmentc5cclog]
Distance of QR
= Area under the line of velocity-time graph from point Q to point R
=Area of the right triangle
$=\frac{1}{2}(Ts)(20ms^{-1})
=50m$

Solving the equation above for T, we get T=5s

We know the gradients of the line PQ and QR are the same (from the fact that the particle moves with uniform deceleration), therefore, we have

$\frac{(40-20)ms^{-1}}{0-t}=\frac{(20-0)ms^{-1}}{((t+5)-t)s}$

Obviously, t=5s.

Hence, this is the answer the second part of the question, that is, the time taken to cover PQ = t = 5s.

And the answer to the third part of the question, time needed to cover PR = (5 + 5)s =10s.

Now, let's go back to the first question.

We're asked to find the distance between the point P and Q.
[attachment=0c5cclog]MAAS Part 3.JPG[/attachmentc5cclog]
Distance of PQ
= Area under the line of velocity-time graph
=Area of the trapezium
$=\frac{1}{2}(5s)(40+20)ms^{-1}
=150m$
Attached Images
 MAAS Part1.JPG (33.9 KB, 956 views) MAAS Part 2.JPG (36.5 KB, 956 views) MAAS Part 3.JPG (38.0 KB, 956 views)

 September 5th, 2012, 07:39 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs Re: A particle moving in a straight line Nice! This is very well explained and demonstrated graphically in a most straightforward fashion!

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