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Sagot :
Answer
Average velocity in the first 2 seconds = 16 ft/s
Explanation
The average value of a function over an interval [a, b] is given as
[tex]\text{Average value of the function = }\frac{1}{b-a}\int ^b_af(x)dx[/tex]The integral is evaluated over the same interval [a, b]
Since we are asked to find the average velocity over the first 2 seconds, we need to first obtain the funcion for th object's velocity.
Velocity = (dh/dt)
h(t)= -16t² + 48t + 120
Velocity = (dh/dt) = -32t + 48
So, we can then find the average velocity over the first 2 seconds, that is, [0, 2]
[tex]\begin{gathered} \text{Average value of the function = }\frac{1}{b-a}\int ^b_af(t)dt \\ a=0,b=2,f(t)=-32t+48 \\ \text{Average Velocity = }\frac{1}{2-0}\int ^2_0(-32t+48)dt \\ =\frac{1}{2}\lbrack-16t^2+48t\rbrack^{2_{}}_0 \\ =\frac{1}{2}\lbrack-16(2^2)+48(2)\rbrack_{} \\ =0.5\lbrack-16(4)+96\rbrack \\ =0.5\lbrack-64+96\rbrack \\ =0.5\lbrack32\rbrack \\ =16\text{ ft/s} \end{gathered}[/tex]Hope this Helps!!!
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