Question

For a particle moving along $v+x$-axis, acceleration is given as $a+x$. Find the position as a function of time? Given that at $t=0,x=1,v=1$.

• A. $x=lnt$
• B. $x=t^2$
• C. $x=e^t$
• D. $x=t^3$

Answer 1

Answer: (C)

$x=e^t$

Answer is C.
It is given that, a = x $\frac{vdv}{dx}=x\Rightarrow \frac{v^2}{2}=\frac{x^2}{2}+C$
Because, $t=0,x=1$ and $v=1$
Therefore, $C=0$ $\Rightarrow v^2=x^2$
v=x but given that $x=1$ when $v=1$
Therefore, $v=x$ $\frac{dx}{dt}=x\Rightarrow \frac{dx}{x}=dt$
$lnx-t+C\Rightarrow 0=0+C\Rightarrow lnx=t$
$x=e^t$
Hence, the position as a function of time is $x=e^t$.