Answer
(a) a =$\frac{g}{3}$
(b) $x|_{t=3s}= 14.7 m$
(c) $m|_{t=3s}= 29.4 g$
Work Step by Step
(a) We have the following given information on differential equation
xg= $x \frac{dv}{dt} + v^{2}$ .......................(1)
Here, the solution of the equation is :
v=at ...........(R.1)
We all know that the acceleration is the first derivates of the velocity
relative to time, So we have :
a=$\frac{dv}{dt}$ ...........(R.2)
So know, We also know that the velocity is the first derivative of position relative to time, So we have the following :
v= $\frac{dx}{dt}$
dx= v dt
Now, Substituting for v from relation (R.1) and integrating we get the following result
dx= at dt
$ ∫_0^x dx= a ∫_0^t t dt $
x - $x_{0} = a \frac{t^{2}}{2}$
Taking the origin at the starting point, then $x_{0}= 0; So we get: $
x = $\frac{1}{2} at^{2}$ .......... (R.3)
2.
Substituting for v from relation (R.1), $\frac{dv}{dt}$ from relation (R.2) and x from relation (R.3) into equation (1) , we get :
$ \frac{1}{2} at^{2}.g = \frac{1}{2} at^{2} .a + a^{2}t^{2}$
=$\frac{3}{2}a^{2}t^{2}$
For nonzero a or t, we multiply by $\frac{2}{3at^{2}}$ So, we get :
$ a=\frac{g}{3}$
(b) We put our value for t= 3.00 s and a into relation (R3) , So we get ;
$x|_{t=3s}=\frac{1}{2}(\frac{9.8}{3}) (3.00^{2}) = 14.7 m$
Thus, $x|_{t=3 s}= 14. 7 m$
(c) We know , that the relation between the mass of raindrop and distance that it has fallen is given by :
m= kx (R.4)
Here k= 2.00 g/m.
So we know to put our value for x|_{t= 3 s} into relation (R.4) , So we get the mass of the raindrop at t= 3.00s as follows :
$m|_{t=3 s} = (2.00 g/m) . (14.7m) = 29.4 g$
Therefore,
$m|_{t=3 s} = 29.4 g$
