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4-1.Newton's Laws of Motion
hard
A force $F = Be^{-Ct}$ acts on a particle whose mass is $m$ and whose velocity is $0$ at $t = 0$. It’s terminal velocity is :
A
$\frac{C}{{mB}}$
B
$\frac{B}{{mC}}$
C
$\frac{{BC}}{m}$
D
$-\frac{B}{{mC}}$
Solution
$F=m a \Rightarrow \quad m a=B e^{-c t}$
$a=\frac{B e^{-c t}}{m}$
$\frac{d v}{d t}=\frac{B e^{-c t}}{m}$
$\int d v=\int \frac{Be^{-c t}}{m} \cdot d t$
$\Rightarrow v(t)=-\frac{B e^{-c t}}{m c}+k$
$v(t)=\frac{B}{m c}\left(1-v^{-c t}\right)$
$k=\frac{B}{m c}$
Standard 11
Physics
