# A balloon rises at a certain rate (in body), What is velocity of balloon after 40 seconds?

A balloon rises vertically from the ground so that its height after $t$ seconds is $h(t) =\frac12t^2+\frac12t$ feet where $t$ is between $0$ and $60$. What is the velocity of the balloon after $40$ seconds?

I could really use some help with this step of the problem. I figured out that the height of the ballon after $40$ seconds is $820$ feet (just $h$ evaluated at $40$). I figured out the average velocity is $20.5~ft/sec$. I now have to determine the velocity after $40$ seconds. I know the formula for inst. velocity. The $t$ would be $40$. The $t+h$ would be $40+h$. But I can't figure out how to get the $f(t)$ and $f(t+h)$ so I can then plug them into the formula where I take the limit of the difference quotient as $h\to0$. This is like $1^{st}$ chapter of a calculus textbook so advanced stuff is not how I must solve this. I cant figure out what function to use to get my final position minus my initial position. The function aboove is for height. I dont know what fcn to use for position? Teaching myself,no professor. Help would be very appreciated. thanks!

• No wind speed was given, so assume that it is zero, in which case, height and position would be synonymous. – John Joy Aug 3 '15 at 18:40

Given that the distance is $h(t)=\frac12t^2+\frac12t$, we need to calculate $h'(40)$ which is the velocity at $40$ seconds. Using the definition of the derivative, we have \begin{align} h'(40)&=\lim_{\Delta t\to 0}\frac{\frac12(40+\Delta t)^2+\frac12(40+\Delta t)-h(40)}{\Delta t}\\ &=\lim_{\Delta t\to 0}\frac{800+40\Delta t+\frac12\Delta t^2+20+\frac12\Delta t-820}{\Delta t}\\ &=\lim_{\Delta t\to 0}\frac{\Delta t(40+\frac12\Delta t+\frac12)}{\Delta t}\\ &=\lim_{\Delta t\to 0}\frac{(40+\frac12\Delta t+\frac12)}{1}=40.5~ft/s.\\ \end{align}

We have, $$h(t)=\frac{1}{2}t^2+\frac{1}{2}t$$ The vertical velocity of balloon $(v)$ will be the rate of change of vertical height $h(t)$ i.e. $\frac{d}{dt}(h(t))$ Hence, by differentiating we get $$\frac{d}{dt}(h(t))=\frac{d}{dt}\left(\frac{1}{2}t^2+\frac{1}{2}t\right)$$ $$v=t+\frac{1}{2}$$ Hence, the vertical velocity at time $t=40 \ sec$, $$v_{t=40}=40+\frac{1}{2}=40.5\ ft/sec$$

The height of the balloon at time $t$ is given by $h(t)=\frac12 t^2+\frac12 t$.

To find the velocity of the balloon, $v(t)$, at time $t$, we can use the definition of the velocity in terms of the limit quotient of the change in height per unit time. That is to find $v(t)$ we evaluate the limit

$$\lim_{\Delta t\to 0}\frac{h(t+\Delta t)-h(t)}{\Delta t}$$

To find $h(t+\Delta t)$ we substitute $t+\Delta t$ for $t$ to find

$$h(t+\Delta t)=\frac12 (t+\Delta t)^2+\frac12 (t+\Delta t)$$

We can then write

\begin{align} v(t)&=\lim_{\Delta t\to 0}\frac{\left(\frac12 (t+\Delta t)^2+\frac12 (t+\Delta t)\right)-\left(\frac12 t^2+\frac12 t\right)}{\Delta t}\\\\ &=\lim_{\Delta t\to 0}\frac{t\Delta t+\frac12\Delta t^2+\frac12 \Delta t}{\Delta t}\\\\ &=t+\frac12 \end{align}

For $t=40$, we find that the velocity $v(40)=40+\frac12$ is

$$\bbox[5px,border:2px solid #C0A000]{v(40)=40.5\,\text{ft/s}}$$

• Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:15