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Physics 15 Online
OpenStudy (anonymous):

The velocity v, in meters per second, is given as a function of time t, in seconds, by v(t)=-0.539t^2+1.35t-7.29. What is the acceleration at time t = 3.21 s? -i took the derivative and that didn't seem to work, i also did it twice to see if i did something wrong, but that didn't work either. please help :)

OpenStudy (abhisar):

v(t)=\(\sf -0.539t^2+1.35t-7.29\\ Acceleration=\frac{dv}{dt}=-1.078t+1.35\\ Plug~in~the~value~of~t\)

OpenStudy (anonymous):

i figured it out last night, but thanks again :)

OpenStudy (anonymous):

i didn't put the negative sign, minor mistake

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