Velocity squared vs position graph

AI Thread Summary
The discussion centers on the interpretation of the slope from a velocity squared versus position graph. The calculated slope, with units of cm/seconds squared, is questioned regarding its relation to acceleration, which is expected to be 9.80 m/sec². Clarification is provided that the slope represents twice the acceleration (2a) when using the equation v² = u² + 2as, where 's' refers to displacement, not time. Confusion arises from the terminology, as 's' is often misinterpreted as seconds. Ultimately, the slope does help determine acceleration when correctly applied in the context of the graph.
azn4lyf89
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I calculated a slope from a best fit line of a velocity squared vs position graph but I don't understand what this slope represents. My units come out to be in cm/seconds squared, so does that mean that's my acceleration? It seems to be way off from 9.80m/sec^2 which should be my acceleration. Or does this slope help me determine to acceleration?
 
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Well from v^2=u^2+2as

If you plot v2 vs. s, your gradient should be 2a
 
rock.freak667 said:
Well from v^2=u^2+2as

If you plot v2 vs. s, your gradient should be 2a

Isn't that only if the graph was velocity squared vs time? Because s is seconds right?
 
azn4lyf89 said:
Isn't that only if the graph was velocity squared vs time? Because s is seconds right?

s is displacement.

t is usually used to inidicate time.
 
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