1. ## Mechanics Question

A glider of mass 200kg needs to travel at 50 m/s before it can take off. It starts from rest and a constant force of 500N pulls it along. If air resistance is $\displaystyle 10v$ N, where $\displaystyle v$ is the speed of the glider, how long does the glider take to get air born?

If the runway needs to be at least 200 m longer than the required take of distance, what should it's minimum length be?
Not sure where to start with this question? Constant force must mean that the acceleration is constant right? So does this mean that I can use the SUVAT equations in my calculations?

Thank in advance for any pointers

Craig

2. Managed to make a start, not sure how correct this is but here's what I've got so far:

Using $\displaystyle F = ma$, we have $\displaystyle 500 - 10v = 250\frac{dv}{dt}$

Rearranging we get the following:

$\displaystyle \int (50 - v)dt = \int 25 dv$, (I think you can do this because $\displaystyle v$ is speed, and therefore a constant?)

$\displaystyle (50 - v)t = 25v$.

Holds true for the initial, $\displaystyle t = 0, v = 0$.

At $\displaystyle v = 40$, $\displaystyle 10t = 1000$, so $\displaystyle t = 100s$.

Is this anywhere near the right method?

3. Ok ignore that last post, realised that you can't just take $\displaystyle v$ as a constant as obviously it's a function of time... (oops)

We have $\displaystyle 500 - 10v = 250\frac{dv}{dt}$, and rearranging we get:

$\displaystyle \frac{dv}{dt} + \frac{1}{25}v = 2$

Using the integrating factor, $\displaystyle e^{\frac{1}{25}t}$, we get:

$\displaystyle \int e^{\frac{1}{25}t}\frac{dv}{dt} + e^{\frac{1}{25}t}\frac{1}{25}v = \int 2e^{\frac{1}{25}t}$

$\displaystyle ve^{\frac{1}{25}t} = 2 \int e^{\frac{1}{25}t}$

$\displaystyle ve^{\frac{1}{25}t} = 50 e^{\frac{1}{25}t} + C$, so $\displaystyle v = 50 + Ce^{\frac{-1}{25}t}$.

Using the initial conditions, we get $\displaystyle C = -50$.

Therefore the equation for the velocity is:

$\displaystyle v = 50 - 50e^{\frac{-1}{25}t}$, and therefore the time at speed $\displaystyle 40m/s$ would be 40.2 seconds to 3sf?

4. yes

but in the original post you say the mass is 200 not 250

so confirm the mass and change the equation if necessary