A body travels uniformly a distance of (13.8 | vedclass

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(B) Given: Distance $s = (13.8 \pm 0.2) 	ext{ m}$ and time $t = (4.0 \pm 0.3) 	ext{ s}$.Velocity $v = \frac{s}{t} = \frac{13.8}{4.0} = 3.45 	ext{ m

Vedclass · Accepted Answer

$(3.5 \pm 0.31) 	ext{ m/s}, \pm 9\%$

Vedclass · Answer

(B) Given: Distance $s = (13.8 \pm 0.2) 	ext{ m}$ and time $t = (4.0 \pm 0.3) 	ext{ s}$.Velocity $v = \frac{s}{t} = \frac{13.8}{4.0} = 3.45 	ext{ m/s} \approx 3.5 	ext{ m/s}$.The relative error in velocity is given by $\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t}$.Substituting the values: $\frac{\Delta v}{v} = \frac{0.2}{13.8} + \frac{0.3}{4.0} = 0.0145 + 0.075 = 0.0895$.Absolute error $\Delta v = v 	imes 0.0895 = 3.45 	imes 0.0895 \approx 0.3087 \approx 0.31 	ext{ m/s}$.Thus,velocity $v = (3.5 \pm 0.31) 	ext{ m/s}$.Percentage error = $\frac{\Delta v}{v} 	imes 100 = 0.0895 	imes 100 = 8.95\% \approx 9\%$.

$A$ body travels uniformly a distance of $(13.8 \pm 0.2) \text{ m}$ in a time $(4.0 \pm 0.3) \text{ s}$. Its velocity with error limits and percentage error is

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