Differential Equations Question 152
Question: A particle moves in a straight line with a velocity given by $ \frac{dx}{dt}=x+1 $ (x is the distance described). The time taken by a particle to traverse a distance of 99 metre is
Options:
A) $ {\log _{10}}e $
B) $ 2{\log _{e}}10 $
C) $ 2{\log _{10}}e $
D) $ \frac{1}{2}{\log _{10}}e $
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Answer:
Correct Answer: B
Solution:
$ \frac{dx}{dt}=x+1 $
Therefore $ \log (x+1)=t+c $
Putting $ t=0,\ x=0 $ , we get $ \log 1=c $
Therefore $ c=0 $
$ \therefore t=\log (x+1) $ . For $ x=99,\ t={\log _{e}}100=2{\log _{e}}10 $ .
BETA
