Question

A body travels uniformly a distance of $(13.7\pm 0.2)m$, in time $(4.0\pm 0.3)s$. The velocity of particle is ?

• A. $(3.45\pm 0.31)m/s$
• B. $(3.4\pm 0.3)m/s$
• C. $(3.68\pm 0.4)m/s$
• D. $(3.6\pm 0.42)m/s$

Answer 1

The correct option is A $(3.45\pm 0.31)m/s$

Given,

$d=(13.7\pm 0.2)m$

$t=(4.0\pm 0.3)s$

$v=\frac{13.7}{4.0}=3.45m/s$

Velocity, $v=\frac{d}{t}$

Take log both side,

$log\left(v\right)=log\left(d\right)-log\left(t\right)$

$\frac{\Delta v}{v}=\frac{\Delta d}{d}+\frac{\Delta t}{t}$

$\frac{\Delta v}{3.45}=\frac{0.2}{13.7}+\frac{0.3}{4.0}=0.0895$

$\Delta v=3.45\times 0.0895=0.31$

Velcity of the particle, $v_p=v\pm \Delta v$

$v_p=(3.45\pm 0.31)m/s$

The correct option is A.