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Distance under constant acceleration |
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| Nov6-06, 06:24 AM | #1 |
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Distance under constant acceleration
A body starts from rest with an acceleration of 11m/s².
What is it's velocity after 6 seconds. (I got 66 for that) and how far did it travel? What forumla do I have to use for this? |
| Nov6-06, 06:43 AM | #2 |
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There is a well-known expression which represents displacement under constant acceleration. Try to find it. (Either here or google it up.)
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| Nov6-06, 06:54 AM | #3 |
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Is displacement the same as distance?
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| Nov6-06, 06:56 AM | #4 |
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Distance under constant acceleration |
| Nov6-06, 07:01 AM | #5 |
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What about the first part of the question?
I tried to average out the velocity and then using the d=vt formula I got 198m. |
| Nov6-06, 07:11 AM | #6 |
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The other formula would be: d = (1/2)at2. Check that the solution is the same. Try to show that v(average)*t = (1/2)at2 |
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