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A particle is moving along a line with velocity function v(t)=t^2-t-12feet per second. Find the displacement on the interval 1is less than or equal to t, and t is less than or equal to 5. And the total distance traveled on the interval
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displacement is the integration of velocity along the interval
so do I derive the function
no, you integrate it
so I plug in 1 and 5 in the equation
im assuming you know what a derivative or an integral already is ....
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\[\int_{a}^{b}v(t)~dt=displacement\]
if you know of some other way to define the area underneath the velocity curve, then by all means have at it
okay so I GOT V(T)DT=(T^3/3-T^2/2-12T)
good, now we evaluate that from t = 1 to 5
distance im am thinking is the length of that curve, from 1 to 5
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