Suppose that while waiting for a big wave, a surfer is being propelled toward the beach in such a way that the distance (in meters) traveled is given by f (t) = 10t2 for 0 ≤ t ≤ 1, where t is in seconds. find the velocity v 1 10 of the surfer at t = 1 10 sec.

your answer is, orbit around the nucleus

there are no correct statementd on the list of choices you provided.

The velocity at t=1/10s will be:

[tex]v(\frac{1}{10})=2\frac{m}{s}[/tex]

Explanation:

The distance in meter as a function of time is:

[tex]f (t) = 10t^2[/tex], for 0 ≤ t ≤ 1 where t is in seconds.

We need to find the velocity [tex]v(\frac{1}{10})[/tex] of the surfer at [tex]t=\frac{1}{10}s[/tex].

For this we can use the definition of instantaneus velocity (in it's limit form):

[tex]v(t) = \lim_{\Delta t \to 0} \frac{f(t+\Delta t)-f(t)}{\Delta t} = \lim_{\Delta t \to 0} \frac{10(t+\Delta t)^2 - 10t^2}{\Delta t} \\= \lim_{\Delta t \to 0} \frac{10(t^2+2t\Delta t+\Delta t^2) - 10t^2}{\Delta t} = \lim_{\Delta t \to 0} \frac{20t\Delta t+10\Delta t^2}{\Delta t} \\= \lim_{\Delta t \to 0} \frac{(20t+10\Delta t)\Delta t}{\Delta t} = \lim_{\Delta t \to 0} 20t+10\Delta t = 20t[/tex]

At [tex]t=\frac{1}{10}s[/tex], we have:

[tex]v(\frac{1}{10}) = 20\frac{1}{10}=2\frac{m}{s}[/tex].