Author Topic: Kinematics (Read 2681 times) Tweet Share

Tea.bag · « **on:** August 19, 2008, 09:45:41 pm »

i cant do this question!!
can someone explain how to do this??

1. a body at rest starts moving with an initial acceleration of 5m/s^2. the acceleration decreases uniformly with the distance travelled reaching a value of zero when the body has travelled 80 metres. Find the maximum speed of the body.

Collin Li · « **Reply #1 on:** August 19, 2008, 09:53:47 pm »

Acceleration decreases uniformly, hence acceleration is a linear function of distance travelled:

Since the body starts at rest, displacement is distance (as there is no change in direction, and acceleration is only decreasing -- no deceleration):

$a = mx + c$

You know that (0,5) initially, and (80, 0) when the acceleration decreases to zero after 80 metres.

Hence: $a = -\frac{1}{16}x + 5$

To find maximum speed of body, we want to find velocity. In order to get a velocity, we must use the fact that: $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$ (the other ones will not work - they do not have the right variables).

Therefore: $\frac{d}{dx}\left(\frac{1}{2}v^2\right) = -\frac{1}{16}x + 5$

$\implies \frac{1}{2}v^2 = -\frac{1}{32}x^2 + 5x + C$

Since the body starts at rest, (0,0), $\implies C = 0$

$\implies \frac{1}{2}v^2 = -\frac{1}{32}x^2 + 5x$

The body stops accelerating at $x=80$ , so this will be its maximum speed:

$\frac{1}{2}v^2 = -\frac{80^2}{32} + 5(80)$

$\implies v = \sqrt{400} = 20 \mbox{ ms}^{-1}$

Mao · « **Reply #2 on:** August 19, 2008, 09:55:10 pm »

$\frac{da}{dx}=-k$

$\implies a=-kx+a_0$

   $a_0=5, k=\frac{5}{80}=\frac{1}{16}$

$\implies \frac{d}{dx}\left(\frac{1}{2}v^2\right)=-\frac{x}{16}+5$

$\implies \frac{1}{2}v^2=-\frac{x^2}{32}+5x+C$

   given that when x=0, v=0, $\implies C=0$

$\implies \frac{1}{2}v^2=-\frac{x^2}{32}+5x$

   since acceleration was positive all the way until displacement is 80, the maximum velocity is reached at x=80

$\implies \frac{1}{2}(v_{max})^2=-\frac{80^2}{32}+400$

$\implies v_{max}=20$

EDIT: oops arithmetic error at last step

Tea.bag · « **Reply #3 on:** August 19, 2008, 09:58:08 pm »

two diffrent answers??
which one is right
btw my book says 20m/s so i dont know if the book is right or not

Collin Li · « **Reply #4 on:** August 19, 2008, 10:00:11 pm »

My one did the linear equation incorrect. I will try to fix it.

Mao · « **Reply #5 on:** August 19, 2008, 10:02:00 pm »

sorry, it is 20. i made an arithmetic error at the last step

Tea.bag · « **Reply #6 on:** August 19, 2008, 10:07:45 pm »

thnx a lot guys...
this whole time i was trying to find velocity in terms of time..i over looked the fact that acceleration can be in terms of position

Tea.bag · « **Reply #7 on:** August 26, 2008, 09:55:46 pm »

another question.

1) A car slows down with a constant retardation from 24m/s to 16m/s over a distance of 15 metres. What further distance will it travel before coming to rest?

Flaming_Arrow · « **Reply #8 on:** August 26, 2008, 10:05:15 pm »

Quote from: Tea.bag on August 26, 2008, 09:55:46 pm

another question.

1) A car slows down with a constant retardation from 24m/s to 16m/s over a distance of 15 metres. What further distance will it travel before coming to rest?

$v = 16 u =24 d=15 a= ?$

$v^2 = u^2 + 2ad$

$256 = 576+ 2*a*15$

$-320 = 30a$

$a = -10.7 ms^{-2}$

$a = -10.7 v = 0 u =16 d= ?$

$v^2 = u^2 + 2ad$

$0 = 256 + 2 * -10.7 * d$

$-256 = -21.4d$

$d = 11.96m$

hence it will travel 12m before it comes to a rest

duno if this is right

shinny · « **Reply #9 on:** August 26, 2008, 10:06:52 pm »

Its...mostly correct. Just dont round off the acceleration (its $-\frac{32}{3}$ ) and you'll get 12 metres exactly.

Tea.bag · « **Reply #10 on:** August 26, 2008, 10:10:40 pm »

thnx..

1 more

Two trains pass one another, travelling in opposite directions on parallel tracks. when the fronts of the trains are in line they are travelling at 12m/s and 16m/s and accelerating at 0.5m/s^2 and 1m/s^2 respectively. the length of each train is 136 metres. how long does it take the rear of the two trains to pass?

/0 · « **Reply #11 on:** August 26, 2008, 10:16:03 pm »

The relative velocity of one train with respect to the other is $12+16 = 28 m/s$ by vector addition. Likewise, as acceleration is a vector, you can see that the acceleration of one train with respect to the other is $1.5m/s^2$

Hence you can use

$x = ut+\frac{1}{2}at^2$

$136 = 28t + 1.5t^2$

$t = 4$ seconds. ( i think )

(yeah, oops, the distance should be twice 136m = 272 m) but hey, that still gives the wrong answer :/ t = 7

Tea.bag · « **Reply #12 on:** August 26, 2008, 10:17:37 pm »

nope the answer says 8 seconds

Flaming_Arrow · « **Reply #13 on:** August 26, 2008, 10:21:18 pm »

Quote from: DivideBy0 on August 26, 2008, 10:16:03 pm

The relative velocity of one train with respect to the other is $12+16 = 28 m/s$ by vector addition. Likewise, as acceleration is a vector, you can see that the acceleration of one train with respect to the other is $1.5m/s^2$

Hence you can use

$x = ut+\frac{1}{2}at^2$

$272 = 28t + 1.5t^2$

$t = 4$ seconds. ( i think )

(yeah, oops, the distance should be twice 136m = 272 m) but hey, that still gives the wrong answer :/

u forgot to divide acceleration by 2

$272 = 28t + 0.75t{^2}$

$t = 8s$

Tea.bag · « **Reply #14 on:** August 26, 2008, 10:25:09 pm »

why do u add the velocities and the acceleration together??

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