Answer :
Answer:
a) h = 0.1: [tex]\bar v = -11\,\frac{ft}{s}[/tex], h = 0.01: [tex]\bar v = -10.1\,\frac{ft}{s}[/tex], h = 0.001: [tex]\bar v = -10\,\frac{ft}{s}[/tex], b) The instantaneous velocity of the ball when [tex]t = 2\,s[/tex] is [tex]-10[/tex] feet per second.
Step-by-step explanation:
a) We know that [tex]y = 30\cdot t -10\cdot t^{2}[/tex] describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ([tex]\bar v[/tex]), measured in feet per second, can be done by means of the following definition:
[tex]\bar v = \frac{y(2+h)-y(2)}{h}[/tex]
Where:
[tex]y(2)[/tex] - Position of the ball evaluated at [tex]t = 2\,s[/tex], measured in feet.
[tex]y(2+h)[/tex] - Position of the ball evaluated at [tex]t =(2+h)\,s[/tex], measured in feet.
[tex]h[/tex] - Change interval, measured in seconds.
Now, we obtained different average velocities by means of different change intervals:
[tex]h = 0.1\,s[/tex]
[tex]y(2) = 30\cdot (2) - 10\cdot (2)^{2}[/tex]
[tex]y (2) = 20\,ft[/tex]
[tex]y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}[/tex]
[tex]y(2.1) = 18.9\,ft[/tex]
[tex]\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}[/tex]
[tex]\bar v = -11\,\frac{ft}{s}[/tex]
[tex]h = 0.01\,s[/tex]
[tex]y(2) = 30\cdot (2) - 10\cdot (2)^{2}[/tex]
[tex]y (2) = 20\,ft[/tex]
[tex]y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}[/tex]
[tex]y(2.01) = 19.899\,ft[/tex]
[tex]\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}[/tex]
[tex]\bar v = -10.1\,\frac{ft}{s}[/tex]
[tex]h = 0.001\,s[/tex]
[tex]y(2) = 30\cdot (2) - 10\cdot (2)^{2}[/tex]
[tex]y (2) = 20\,ft[/tex]
[tex]y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}[/tex]
[tex]y(2.001) = 19.99\,ft[/tex]
[tex]\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}[/tex]
[tex]\bar v = -10\,\frac{ft}{s}[/tex]
b) The instantaneous velocity when [tex]t = 2\,s[/tex] can be obtained by using the following limit:
[tex]v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}[/tex]
[tex]v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}[/tex]
[tex]v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}[/tex]
[tex]v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}[/tex]
[tex]v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}[/tex]
[tex]v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h[/tex]
[tex]v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h[/tex]
[tex]v(t) = 30-20\cdot t[/tex]
And we finally evaluate the instantaneous velocity at [tex]t = 2\,s[/tex]:
[tex]v(2) = 30-20\cdot (2)[/tex]
[tex]v(2) = -10\,\frac{ft}{s}[/tex]
The instantaneous velocity of the ball when [tex]t = 2\,s[/tex] is [tex]-10[/tex] feet per second.