Distance under constant acceleration

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A body starting from rest with an acceleration of 11 m/s² reaches a velocity of 66 m/s after 6 seconds. To calculate the distance traveled, the formula d = (1/2)at² can be used, which gives a displacement of 198 meters. Displacement is considered the same as distance in this context. The average velocity method also confirms this result, as it shows that v(average) multiplied by time equals the displacement. Both approaches validate the calculations under constant acceleration.
Alpha[X]²
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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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There is a well-known expression which represents displacement under constant acceleration. Try to find it. (Either here or google it up.)
 
Is displacement the same as distance?
 
Alpha[X]² said:
Is displacement the same as distance?

Yes it is, in this case.
 
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.
 
Alpha[X]² said:
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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