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how do you derive velocity from the equation 86t+t^2? PLEASE EXPLAIN HOW YOU GOT THE ANSWER. Thanks.
 2 years ago
 2 years ago
how do you derive velocity from the equation 86t+t^2? PLEASE EXPLAIN HOW YOU GOT THE ANSWER. Thanks.
 2 years ago
 2 years ago

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mohammadhBest ResponseYou've already chosen the best response.2
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^(21). 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^(11) +(1*2)*t^(21) = 0 6*t^0 + 2*t^1 = 06+2t
 2 years ago

study122Best ResponseYou've already chosen the best response.0
since the acceleration function has already been giving as a=86t+t^2,we want to find velocity,it implies we have to integrate i.e a=86t+t^2 implies \[v=\int\limits a dt = \int\limits (86t+t^2)dt\] . 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.
 2 years ago
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