Question

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

• A. $e-1$
• B. $e^2$$-1$
• C. $\frac{e}{2}$
• D. $e+1$

Answer 1

The correct option is A $e-1$

Answer is A.
In this case, $a=v$$\frac{dv}{dt}=v$
$lnv=t+c0=0+c$
$v=e^{\prime }$$\frac{dx}{dt}=e^{\prime }$
That is, $x=e^{\prime }+c0=1+c$
$x=e^{\prime }-1$
Hence, the position as a function of time is $e-1$.