Question

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.

Answer 1

Answer:

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

$v(\frac{1}{10})=2\frac{m}{s}$

Explanation:

The distance in meter as a function of time is:

$f (t) = 10t^2$, for 0 ≤ t ≤ 1 where t is in seconds.

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

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

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

At $t=\frac{1}{10}s$, we have:

$v(\frac{1}{10}) =  20\frac{1}{10}=2\frac{m}{s}$.