Sep 29, 2017

Velocity $v \left(m {s}^{-} 1\right)$ satisfies $3.16 \le v \le 3.78$ and b) is the best answer.

#### Explanation:

Calculating the upper and lower bound helps you in this type of problem.

If the body travels the longest distance ($14.0 m$) in the shortest
time ($3.7 s$), the velocity is maximized. This is the upper bound
of the velocity ${v}_{\max}$

${v}_{\max}$ = $\frac{14.0 \left(m\right)}{3.7 \left(s\right)}$ = $3.78 \left(m {s}^{-} 1\right)$.

Simirally, the lower bound of the velocity ${v}_{\min}$ is obtained as
${v}_{\min}$ = $\frac{13.6 \left(m\right)}{4.3 \left(s\right)}$ = $3.16 \left(m {s}^{-} 1\right)$.

Therefore, the velocity $v$ stands between $3.16 \left(m {s}^{-} 1\right)$ and $3.78 \left(m {s}^{-} 1\right)$. Choice b) fits this best.

Sep 29, 2017

Option (b)
$\left(3.45 \pm 0.30\right) \frac{m}{s}$

#### Explanation:

if the Quantity is defined as $x = \frac{a}{b}$
let $\Delta a = \text{Absolute error for a}$
$\Delta b = \text{Absolute error for b}$
$\Delta x = \text{Absolute error for x}$
then The Maximum Possible Relative Error in x is
$\frac{\Delta x}{x} = \pm \left[\frac{\Delta a}{a} + \frac{\Delta b}{b}\right]$

Now
Distance $= \left(13.8 \pm 0.2\right) m$
$s = 13.8 m$ and $\Delta s = 0.2 m$
Time $= \left(4.0 \pm 0.3\right) m$
$t = 4.0 m$ and $\Delta t = 0.3 m$

Velocity of Body Within error Limit is $v + \Delta v$
Now $\text{velocity"="Distance"/"time}$
$v = \frac{s}{t} = \frac{13.8}{4} = 3.45 \frac{m}{s}$

and Relative error in Velocity
$\frac{\Delta v}{v} = \pm \left[\frac{\Delta s}{s} + \frac{\Delta t}{t}\right]$
$\frac{\Delta v}{v} = \pm \left[\frac{0.2}{13.8} + \frac{0.3}{4}\right] = 0.014 + 0.075 = 0.089$
Absolute Error in Velocity
$\Delta v = 0.089 \times v = 0.089 \times 3.45 = 0.307 \frac{m}{s}$

Hence

Velocity of Body Within error Limit is
$v + \Delta v = \left(3.45 \pm 0.30\right) \frac{m}{s}$

Option (b)