how do you derive velocity from the equation 8-6t+t^2? PLEASE EXPLAIN HOW YOU GOT THE ANSWER. Thanks.
MIT 8.01 Physics I Classical Mechanics, Fall 1999
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hello, i'm not quit very good in math, but i'll give it a go.
this is a general case: the derivative of a constant is zero
whereas, for example, the derivative of (3x)^2, where x is a variable, is (3*2)*x^(2-1). we multiply the exponent of the term (3x) by the constant next to the variable, and we subtract 1 from the exponent.
in case of the question: derivation:0- (6*1)*t^(1-1) +(1*2)*t^(2-1) = 0- 6*t^0 + 2*t^1 = 0-6+2t
since the acceleration function has already been giving as a=8-6t+t^2,we want to find velocity,it implies we have to integrate
i.e a=8-6t+t^2 implies \[v=\int\limits a dt = \int\limits (8-6t+t^2)dt\] . after integration this is what you get\[c + 8t -3t^2 +t^3/3\] which becomes your velocity.
now if you know your initial velocity and time that is V1 and T1 respectively you substitute into your velocity equation \[v =c + 8t -3t^2 +t^3/3\] you can derive your constant "C".
hence to find the second velocity all you have to do is substitute the time T2 of the second velocity into the equation.