1I was taking a test
cant understand this one!!
if the acceleration is directly proportional to time. then the displacement is?
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5 Answers
49a \propto t \Rightarrow a=kt \Rightarrow \frac{dv}{dt}=kt \Rightarrow \int_{v_o}^{v}{dv}=k\int_{0}^{t}{tdt}\Rightarrow v=v_o+k\frac{t^2}{2}
\Rightarrow \frac{dx}{dt}=v_o+k\frac{t^2}{2}\Rightarrow \int_{x_o}^{x}{dx}=\int_{0}^{t}\left[ {v_o+k\frac{t^2}{2}}\right]dt\Rightarrow x=x_o+v_ot+k\frac{t^3}{6}
k= constant
vo=velocity at t=0
xo=displacement at t=0
49yea now i think these are easy....
never thought so when i was in 11 ka starting! :D
1a∞t
or a=kt where k is the proportionality constant...
so v=∫kt=k∫t
or v=kt2/2
or s=∫kt2/2
or s=k/2∫t2
or s=k/2*t3/3
so it can be determined but it is neither zero nor constant...