Question

The speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is given by $V=3\sqrt{12-x^2}$. All quantities are in SI units.

• A. The motion is uniformly accelerated along the straight line
• B. The magnitude of acceleration at a distance 3 m from the fixed point is $27m{s}^{-2}$
• C. The magnitude of acceleration at a distance 3 m from the fixed point is $9m{s}^{-2}$
• D. The maximum displacement of the particle from the fixed point is 4 m

Answer 1

Answer: (B)

The magnitude of acceleration at a distance 3 m from the fixed point is $27m{s}^{-2}$

$V=3\sqrt{12-x^2}$
Thus, $a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}$ by chain rule
or, $a=v\frac{dv}{dx}=\left(3\sqrt{12-x^2}\right).(-6x).1/2.(12-x^2{)}^{-1/2}$
At x = 3, $a=3.\sqrt{3}.(-18)/2.\frac{1}{\sqrt{3}}=-27m/s^2$
Thus magnitude = 27, hence B is correct.
A is wrong because acceleration is not uniform, but dependent on x.
D is incorrect because maximum displacement means v=0, which gives 0=$12-x^2$ or $x=\sqrt{12}$