Derivative with respect to time ?

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To determine how long it takes for a marble's velocity to reduce from 1.72 m/s to 0.86 m/s in a liquid where acceleration is given by a = -4.8v^2, the discussion emphasizes the need to take the derivative with respect to time. The user initially struggles with this concept but receives guidance on setting up the equation with a = (dv/dt) and rearranging it to integrate. The integration leads to the formulation (dv/(v^2)) = -4.8dt, which allows for solving the time variable. This method effectively connects the velocity and time through calculus. The conversation concludes with the user expressing satisfaction after understanding the integration process.
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derivative with respect to time ??

ok i got a question that reads the accel of a marble is a certain liquid is given by the following equation: a=§ v^2
where §=-4.8 and v> 0m/s the initial volociy is 1.72 m/s. how long will it take the marble's volocity to be reduced to half of it's initial which is 0.86 m/s?

i think i have to take a derivative with repect to time to get the volocity equation and kinamatics doesn't work i can't just find the accel with the equation because the accel has an exponetial relation to volocity...i need hlep. i need to know how to take a derivative with repect to time, i can do derivative fine but with repect to time i;m stumped. HELP ! :confused:
 
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do you know the answer?
 
nm i got it dv/dt then integrate
 
Try this a=(dv/dt) and put this in a=-4,8v^2. Then write all v's to one side and the t's to the other to get : (dv/(v^2))=-4,8dt and start integrating.

regards
marlon
 
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