The position of a softball tossed vertically upward is described by the equation y=c1(t)−c2(t^2), wherey is in meters,tin seconds,c1=6.71 m/s, and c2= 2.39 m/s^2. How would you find the acceleration at a specific t value?
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\[\Large y=c_1t-c_2t^2\] `Position` is usually denoted by s as a function of t:\[\Large s(t)=c_1t-c_2t^2\]
The instantaneous rate of change of postion gives us velocity,\[\Large s'(t)=v(t)\]The second derivative of position,
or the first derivative of velocity,
gives us acceleration.\[\Large s''(t)=a(t)\]
Then to find the acceleration at a specific \(\Large t\) value, let's say at \(\Large b\), we would plug that value into the acceleration function.
So it looks like you'll want to find the second derivative of your function :o
Understand how to do that?
I did that and got 4.78 but apparently that is wrong