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November 01, 2025, 07:54:02 pm

Author Topic: Kinematics help  (Read 940 times)  Share 

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Chavi

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Kinematics help
« on: July 28, 2010, 05:28:38 pm »
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Hey forum,

I'm wondering if I can get some help with the following kinematics question:


1. A particle is brought to top speed with an accel which varies linearly as the distance travelled. It starts from rest with an accel of 3 m/s^2 and reaches top speed in a distance of 160m. Find:
a) the top speed
b) the speed when the particle has moved 80m


Thanks.
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98.40_for_sure

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Re: Kinematics help
« Reply #1 on: July 28, 2010, 05:56:04 pm »
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Do you have the answers? I need something to check with, before i post up how i did it
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Martoman

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Re: Kinematics help
« Reply #2 on: July 28, 2010, 08:35:40 pm »
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First one should be

Second :P

« Last Edit: July 28, 2010, 08:37:21 pm by Martoman »
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Chavi

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Re: Kinematics help
« Reply #3 on: July 28, 2010, 08:37:03 pm »
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First one should be

Second :P



According to the answers
a) 4sqrt(30) m/s
b) 6sqrt(10) m/s
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Martoman

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Re: Kinematics help
« Reply #4 on: July 28, 2010, 08:53:30 pm »
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The way I got to the first one was:

let T be some time in the future at top speed x.

ie: a pretty triangle.



Also gradient of line is 3.







So x, top speed, = 320/10.33 = 30.98...


this assumes constant acceleration RTFQ
« Last Edit: July 28, 2010, 09:17:01 pm by Martoman »
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Chavi

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Re: Kinematics help
« Reply #5 on: July 28, 2010, 09:02:50 pm »
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my workings thus far have been:
when
so



so find that c =0, and:


now, how do you find k?
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Martoman

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Re: Kinematics help
« Reply #6 on: July 28, 2010, 09:16:15 pm »
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Ok.

Simply you have

When x = 160, a = 0 as it accelerates no more!



Shove it in it works out to your answer for 1.

Then you have your v equation.

Shove in 80 for part 2.

Works out nicely.
« Last Edit: July 28, 2010, 09:18:27 pm by Martoman »
2009: Math methods: 50, Psychology: 44
2010: chem 47, further 48, Spesh 49 fml seriously and other yr 11 subs.
2011: Holidaying, screw school.
No. Not azn.
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Chavi

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Re: Kinematics help
« Reply #7 on: July 28, 2010, 09:21:25 pm »
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Ok.

Simply you have

When x = 160, a = 0 as it accelerates no more!



Shove it in it works out to your answer for 1.

Then you have your v equation.

Shove in 80 for part 2.

Works out nicely.
Thanks - i was experimenting with diff values - just overlooked the correct figure.
2009: Math Methods CAS [48]
2010: English [47]|Specialist Maths[44]|Physics[42]|Hebrew[37]|Accounting[48]  atar: 99.80
My blog: http://diasporism.wordpress.com/