Author Topic: Kinematics (Read 13562 times) Tweet Share

d0minicz · « **on:** July 04, 2009, 06:08:03 pm »

A particle moving in a straight line passes through a fixed point O, with a velocity of 30 m/s. The acceleration $a m/s^2$ , of the particle, t seconds after passing O, is given by $a= 13-6t$ . Find:
a) the velocity of the particle three seconds after passing O
b) the time taken to reach the maximum distance from O in the direction of the intial moion
c) the value of this maximum distance

thanks.

Flaming_Arrow · « **Reply #1 on:** July 04, 2009, 06:14:34 pm »

$\frac{dS}{dt}= \int 13-6t dt = 13t-3t^2+c$

$30=c \mbox{when t=0}$

$\frac{dS}{dt} = 13t-3t^2+30$

then just sub in t = 3 for part a

Damo17 · « **Reply #2 on:** July 04, 2009, 06:20:39 pm »

Quote from: d0minicz on July 04, 2009, 06:08:03 pm

A particle moving in a straight line passes through a fixed point O, with a velocity of 30 m/s. The acceleration $a m/s^2$ , of the particle, t seconds after passing O, is given by $a= 13-6t$ . Find:
a) the velocity of the particle three seconds after passing O
b) the time taken to reach the maximum distance from O in the direction of the intial moion
c) the value of this maximum distance

thanks.

a) $v= \int a \ dt =13t-3t^2 + 30$ ( $c=30$ )

$v(3)= 13(3)-3(9)+30=42ms^{-1}$

b) $v=0$ , $0=13t-3t^2+30$ , $0= -(x-6)(3x+5)$ , $\therefore t=6s$

c) $x= \int v \ dt = \frac{13t^2}{2} - t^3 +30t$ , (d=0 as when t=0 x=0)

$x(6)=198m$

d0minicz · « **Reply #3 on:** July 04, 2009, 06:46:24 pm »

An object travels in a line such that its velocity v m/s at time t seconds is given by $v= cos(\frac {t}{2}) , t \in [0,4\pi]$ . Given that the initial position of the object is 0.5:
a) find an expression for the position x of the object in terms of t
b) sketch the position-time graph
c) find an expression for the acceleration of the object in terms of t
d)Find an expression (excluding t) for:
i) displacement in terms of acceleration
ii) displacement in terms of velocity
iii) velocity in terms of acceleration

thanks in advance guys

kamil9876 · « **Reply #4 on:** July 04, 2009, 07:06:52 pm »

Not sure if this is the only answer but:
$<br />x=2sin(t/2)+0.5$
$=2\sqrt{1-cos^2(t/2)}*sign(x)+0.5$
$=2\sqrt{1-v^2}*sign(2sin(t/2))+0.5$

So you can basically break it up into a hybrid function:

$x=2\sqrt{1-v^2}+0.5$ for t such that $2\pi k<\frac{t}{2}<2\pi k+\pi$
$x=-2\sqrt{1-v^2}+0.5$ for t such that $(2k-1)\pi<\frac{t}{2}<2k\pi$
Where k is any integer

Note: you should simplify the domains algebraically for simplicity's sake.

Edit: oh yes, don't forget the constant 0.5 at the end.

Damo17 · « **Reply #5 on:** July 04, 2009, 07:17:58 pm »

Quote from: d0minicz on July 04, 2009, 06:46:24 pm

An object travels in a line such that its velocity v m/s at time t seconds is given by $v= cos(\frac {t}{2}) , t \in [0,4\pi]$ . Given that the initial position of the object is 0.5:
a) find an expression for the position x of the object in terms of t
b) sketch the position-time graph
c) find an expression for the acceleration of the object in terms of t
d)Find an expression (excluding t) for:
i) displacement in terms of acceleration
ii) displacement in terms of velocity
iii) velocity in terms of acceleration

thanks in advance guys

d) ii)

$x= 2sin (\frac{1}{2} t) +0.5$

$v^2= cos^2 (\frac{1}{2} t)$

$sin^2 (\frac{1}{2} t) + cos^2 (\frac{1}{2} t)= 1$

$sin (\frac{1}{2} t) = \sqrt{1- cos^2 (\frac{1}{2} t)} = \sqrt{1-v^2}$

so $x= 2\sqrt{1-v^2} + 0.5$

iii) $v= cos (\frac{1}{2} t)$

$sin (\frac{1}{2} t)=-2a, \therefore sin^2 (\frac{1}{2} t) = 4a^2$

$1-sin^2 (\frac{1}{2} t)= cos^2 (\frac{1}{2} t)$

$\therefore \sqrt{1-4a^2}=cos(\frac{1}{2} t) = v$

d0minicz · « **Reply #6 on:** July 04, 2009, 07:22:41 pm »

thx guys

A particle moves horizontally in a line so that its position , x m, from O at time t seconds is given by $x= t^3 - \frac {15}{2} t^2 +12t + 10$ . Find:
~~a) when and where the particle has zero velocity~~
b) the average velocity during the third second
~~c) the velocity at t=2~~
~~d) the distance travelled in the first two seconds~~
e) the closest the particle comes to O.

cheers =]

kamil9876 · « **Reply #7 on:** July 04, 2009, 07:24:26 pm »

Quote from: Damo17 on July 04, 2009, 07:17:58 pm

Quote from: d0minicz on July 04, 2009, 06:46:24 pm
An object travels in a line such that its velocity v m/s at time t seconds is given by $v= cos(\frac {t}{2}) , t \in [0,4\pi]$ . Given that the initial position of the object is 0.5:
a) find an expression for the position x of the object in terms of t
b) sketch the position-time graph
c) find an expression for the acceleration of the object in terms of t
d)Find an expression (excluding t) for:
i) displacement in terms of acceleration
ii) displacement in terms of velocity
iii) velocity in terms of acceleration

thanks in advance guys

d) ii)

$x= 2sin (\frac{1}{2} t) +0.5$

$v^2= cos^2 (\frac{1}{2} t)$

$sin^2 (\frac{1}{2} t) + cos^2 (\frac{1}{2} t)= 1$

$sin (\frac{1}{2} t) = \sqrt{1- cos^2 (\frac{1}{2} t)} = \sqrt{1-v^2}$

so $x= 2\sqrt{1-v^2} + 0.5$

iii) $v= cos (\frac{1}{2} t)$

$sin (\frac{1}{2} t)=-2a, \therefore sin^2 (\frac{1}{2} t) = 4a^2$

$1-sin^2 (\frac{1}{2} t)= cos^2 (\frac{1}{2} t)$

$\therefore \sqrt{1-4a^2}=cos(\frac{1}{2} t) = v$

Yes but don't forget the negative in from the square root that occurs when t is between 2pi and 4pi.

kamil9876 · « **Reply #8 on:** July 04, 2009, 07:26:05 pm »

Quote from: d0minicz on July 04, 2009, 07:22:41 pm

thx guys

A particle moves horizontally in a line so that its position , x m, from O at time t seconds is given by $x= t^3 - \frac {15}{2} t^2 +12t + 10$ . Find:
~~a) when and where the particle has zero velocity~~
b) the average velocity during the third second
~~c) the velocity at t=2~~
~~d) the distance travelled in the first two seconds~~
e) the closest the particle comes to O.

cheers =]

Draw an x versus t graph. You can see that you are after either the minimum positive value of x or the maximum negative. Find those two and then see which has the smaller absolute value

d0minicz · « **Reply #9 on:** July 04, 2009, 07:30:29 pm »

For which part would i do that for ?

kamil9876 · « **Reply #10 on:** July 04, 2009, 07:56:45 pm »

part e. Sorry I didn't see the one sandwhiched between the cross outs lol.

b.) 3rd second means on the interval (2,3). ie: (0,1) is first second, (1,2) is second, (2,3) is third.

The average velocity is just $\frac{\Delta x}{\Delta t}=\frac{x(3)-x(2)}{3-2}$

PS:instantaenous velocity is the limit of that as $\Delta t \rightarrow 0$ ie: the derivative

d0minicz · « **Reply #11 on:** July 04, 2009, 08:38:12 pm »

Oh thanks
Where did (2,3) come from? and how do you know to use it like that?

zzdfa · « **Reply #12 on:** July 04, 2009, 09:03:33 pm »

(0,1) =first second
(1,2) =2nd second
(2,3) =3rd second
average velocity: change in distance over change in time
i hope i understood your question =/

d0minicz · « **Reply #13 on:** July 05, 2009, 01:33:38 am »

hmm im still stuck on those two parts (b and e) sorry guys my fault

d0minicz · « **Reply #14 on:** July 05, 2009, 10:58:08 am »

So say if its the 2nd second
then it'll be
$\frac{x(2) - x (1)} {2-1}$ ?

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