Answer
a) $24.5\;\rm m/s,\;34.6\;m/s$
b) $ a_x= \sqrt{\dfrac{ P }{2mt} } $
c) $3.87\;\rm m/s^2,\;1.22\;m/s^2$
d) It gives unreasonable values.
Work Step by Step
a)
Since the car accelerates at the maximum possible rate during the first 20 s is modeled by
$$v_x^2=\dfrac{2Pt}{m}$$
Thus, the speed of the car from 0 s to 20 s is given by
$$v_x =\sqrt{\dfrac{2\cdot 3.6\times10^4 t}{1200}}$$
$$v_x=\sqrt{60t}\tag 1$$
Thus, at $t=10\;\rm s$, the speed is
$$v_x =\sqrt{60\cdot 10}=\color {red}{\bf24.5}\;\rm m/s$$
And at $t=20\;\rm s$, the speed is
$$v_x =\sqrt{60\cdot 20}=\color {red}{\bf 34.6}\;\rm m/s$$
b) To find the acceleration of the car in terms of $P$, $m$, and $t$, we need to differentiate the original formula relative to $t$.
To make the derivative easy, we need to take $t$ out the square root since all other variables are constants.
$$v_x =\sqrt{\dfrac{2P}{m}}\cdot t^{\frac{1}{2}}$$
$$\dfrac{d}{dt}v_x =\dfrac{d}{dt}\left(\sqrt{\dfrac{2P }{m} }\cdot t^{\frac{1}{2}}\right) $$
$$a_x=\left(\sqrt{\dfrac{2P }{m}}\right) \dfrac{d}{dt} \cdot t^{\frac{1}{2}}=\sqrt{\dfrac{2P }{m}}\cdot \frac{1}{2}t^{\frac{-1}{2}}$$
Thus,
$$\boxed{a_x= \sqrt{\dfrac{ P }{2mt} } } $$
c) The acceleration at $t=1\;\rm s$;
$$a_x= \sqrt{\dfrac{ P }{2mt} } = \sqrt{\dfrac{ 3.6\times10^4}{2\cdot 1200\cdot 1} }=\color{red}{\bf 3.87}\;\rm m/s^2$$
The acceleration at $t=10\;\rm s$;
$$a_x= \sqrt{\dfrac{ P }{2mt} } = \sqrt{\dfrac{ 3.6\times10^4}{2\cdot 1200\cdot 10} }=\color{red}{\bf 1.22}\;\rm m/s^2$$
d) It fails because the smaller the $t$, the huge the acceleration.
Let's take at $t=0.01\;\rm s$ to see the magnitude of the acceleration.
$$a_x= \sqrt{\dfrac{ P }{2mt} } = \sqrt{\dfrac{ 3.6\times10^4}{2\cdot 1200\cdot 0.01} }=\color{blue}{\bf 38.7}\;\rm m/s^2$$
which is unreasonable value.
