A body travels uniformly a distance of (13.8 | vedclass

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(B) Given: Distance $S = (13.8 \pm 0.2) \ m$ and time $t = (4.0 \pm 0.3) \ s$.Velocity $v = \frac{S}{t} = \frac{13.8}{4.0} = 3.45 \ ms^{-1}$.The relat

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$(3.45 \pm 0.3) \ ms^{-1}$

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(B) Given: Distance $S = (13.8 \pm 0.2) \ m$ and time $t = (4.0 \pm 0.3) \ s$.Velocity $v = \frac{S}{t} = \frac{13.8}{4.0} = 3.45 \ ms^{-1}$.The relative error in velocity is given by $\frac{\Delta v}{v} = \frac{\Delta S}{S} + \frac{\Delta t}{t}$.Substituting the values: $\frac{\Delta v}{3.45} = \frac{0.2}{13.8} + \frac{0.3}{4.0}$.$\frac{\Delta v}{3.45} = 0.0145 + 0.075 = 0.0895$.$\Delta v = 3.45 	imes 0.0895 \approx 0.3087 \ ms^{-1}$.Rounding to one significant figure,$\Delta v \approx 0.3 \ ms^{-1}$.Thus,the velocity is $(3.45 \pm 0.3) \ ms^{-1}$.

$A$ body travels uniformly a distance of $(13.8 \pm 0.2) \ m$ in a time $(4.0 \pm 0.3) \ s$. The velocity of the body within error limits is

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