Suppose that an object is at position s(t)= sqrt(t) feet at time t seconds.
A.) Find the average velocity of the object over a time interval from time 4+h seconds to time 4 seconds.
B.) Find the instantaneous velocity of the object at time 4 seconds by taking the limit in part A as h --> 0. (Note, 4 + h--> 4 as h -->0)
+583 the average velocity of the object over a time interval from time 4+h seconds to time 4 seconds is
\(\frac{S(4+h）-S(4)}{(h+4)-4}=\frac{（\sqrt{4+h}-\sqrt{4})}{h}=\frac{（\sqrt{4+h}-\sqrt{4})}{h}*\frac{\sqrt{4+h}+\sqrt{4}}{\sqrt{4+h}+\sqrt{4}}=\frac{h}{h*(\sqrt{4+h}+\sqrt{4})}=1/(\sqrt{4+h}+\sqrt{4})\)
(2)\(\lim_{h\rightarrow 0} 1/(\sqrt{4+h}+\sqrt{4})=1/(\sqrt{4}+\sqrt{4})=1/2\)
