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Distance under constant acceleration

 
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Nov6-06, 06:24 AM   #1
 

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?
 
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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.)
 
Nov6-06, 06:54 AM   #3
 
Is displacement the same as distance?
 
Nov6-06, 06:56 AM   #4
 
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Distance under constant acceleration


Quote by Alpha[X]²
Is displacement the same as distance?
Yes it is, in this case.
 
Nov6-06, 07:01 AM   #5
 
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
 
Quote by Alpha[X]²
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.
Yes, you did it correctly. Remember that you could calculate the displacement averaging out the velocity only because the acceleration is constant.

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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