Calculate Velocity of Particle at t=0.657s

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The particle's position is defined by the equation z = 2.12m/s²t² - 2.96m/s³t³, and the goal is to find its velocity at t = 0.657s. The correct method to find velocity is to take the derivative of the position function with respect to time, rather than using equations that assume constant acceleration. Initial attempts to calculate velocity using displacement over time or the average velocity formula were incorrect due to the non-constant acceleration of the particle. The correct velocity at the specified time is -1.05 m/s, highlighting the importance of using the derivative for accurate results. Understanding the relationship between position and velocity is crucial in solving such problems.
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1. A particle moves in such a way that its position z, in meters, is given as a function of t, in seconds, by the equation z = 2.12m/s^2t^2 - 2.96m/s^3t^3

What is the velocity (m/s) of the particle at t = 0.657s

The answer is -1.05, which is not what I got.

2. First, I used the equation Xf = Xi + 1/2 (Vi + Vf) (t)

Plugging in .0757 for Xf

My answer was V=.231

Then I tried V= Displacement/time

.0757/.657 = .115

What am I doing wrong here?
 
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You are using the wrong equation.

Xf = Xi + 1/2 (Vi + Vf) (t) is valid only when the acceleration is constant. Here it is not. How do you find the velocity as a function of time if you know the position as a function of time? Hint: Velocity is the rate of change of position with respect to time.
 
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I already tried the equation

V=Displacement/time

.0757/.657 = .115

Do I now have to substitute that answer into the equation for position as a function of time?
 
or...do I have to take the derivative of v=displacement/time
 
Yes, you have to take the derivative. v = dz/dt.
 
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