Question 10: Units , Dimensions and Error Analysis

Question

A particle starts from rest and moves along the x -axis with a velocity that varies as $\mathrm{v}=\sqrt{100+4 x} \mathrm{~ms}^{-1}$. The acceleration of the particle is:

Accepted Answer

We can use the chainÃ¢ÂÂrule form of acceleration:
a = dv/dt = (dv/dx)ÃÂ·(dx/dt) = vÃÂ·(dv/dx).
Start with
$$v = \sqrt{100 + 4x}\,. $$
Differentiate w.r.t. x:
$$\frac{dv}{dx} = \frac{1}{2\sqrt{100 + 4x}}\cdot 4 = \frac{2}{\sqrt{100 + 4x}}\,. $$
Multiply by v to get a:
$$a = v\,\frac{dv}{dx}         = \sqrt{100 + 4x}\;\frac{2}{\sqrt{100 + 4x}}        = 2\;	ext{m/s}^2\,. $$
Answer: Option D (2 m/sÃÂ²)

Answer

$25 \mathrm{~ms}^{-2}$

Answer

$5 \mathrm{~ms}^{-2}$

Answer

$4 \mathrm{~ms}^{-2}$

Answer

$2 \mathrm{~ms}^{-2}$

Units , Dimensions and Error Analysis

A particle starts from rest and moves along the x -axis with a velocity that varies as \mathrm{v}=\sqrt{100+4 x} \mathrm{~ms}^{-1}. The acceleration of the particle is:

Choose the correct answer: