# How do you find the instantaneous velocity for the particle whose position at time #t# is given by #s(t)=3t^2+5t# ?

by the derivative of the function

we know that the derivative is a rate of change of a function and that velocity is the rate of change of position ,therefore, if we know the expression for position, its derivative will be its velocity therefore,

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To find the instantaneous velocity of a particle given its position function ( s(t) = 3t^2 + 5t ), you need to find the derivative of the position function with respect to time. The derivative of the position function ( s(t) ) with respect to time ( t ) will give you the instantaneous velocity function ( v(t) ). So, differentiate ( s(t) ) with respect to ( t ) to find ( v(t) ).

( s(t) = 3t^2 + 5t ) ( v(t) = \frac{ds}{dt} ) ( v(t) = \frac{d}{dt}(3t^2 + 5t) ) ( v(t) = 6t + 5 )

Therefore, the instantaneous velocity of the particle at any time ( t ) is given by ( v(t) = 6t + 5 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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