Question

A particle starts moving with initial velocity 3 m/s along x – axis from origin. Its acceleration is varying with x in parabolic nature as shown in figure. At x =$\sqrt{3}$m tangent to the graph makes an angle ${60}^{\circ }$ with positive x – axis as shown in diagram. Then at x = $\sqrt{3}$

• A. $v=\sqrt{(\sqrt{3}+a)}m/s$
• B. $a=1.5m{s}^2$
• C. $v=\sqrt{12}m/s$
• D. $a=3m/s^2$

Answer 1

Answer: (A), (B)

The correct options are
A $v=\sqrt{(\sqrt{3}+a)}m/s$
B $a=1.5m{s}^2$

$a=k{x}^2\frac{da}{dx}=2kx\Rightarrow tan{60}^{\circ }=2k\sqrt{3}\Rightarrow k=\frac{1}{2}a=\frac{x^2}{2}v\frac{dv}{dx}=\frac{x^2}{2}\Rightarrow \frac{v^2}{2}=\frac{x^3}{6}+C_{1}atx=0,v=3m/s{C}_1=\frac{9}{2}{v}^2=\frac{x^3}{3}+9Hencea=1.5andv=\sqrt{\sqrt{3}+9}$