A body travels uniformly a distance of (13.8 | vedclass

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Question

$ s = (13.8 \pm 0.2) m$ 

$ t = (4.0 \pm 0.3) s$

$V  =  \frac{s}{t}\,\, = \,\,\frac{{13.8}}{{4.0}}\,\, = \,\,3.45\,\,m{s^{ - 1}}\,

Vedclass · Accepted Answer

$(3.5 \pm 0.31), \pm 9\%$

Vedclass · Answer

$ s = (13.8 \pm 0.2) m$ 

$ t = (4.0 \pm 0.3) s$

$V  =  \frac{s}{t}\,\, = \,\,\frac{{13.8}}{{4.0}}\,\, = \,\,3.45\,\,m{s^{ - 1}}\,\, = \,\,3.5\,\,m{s^{ - 1}}$

$\frac{{\Delta v}}{v}\,\, = \,\, \pm \,\,\left( {\frac{{\Delta s}}{s}\,\, + \,\,\frac{{\Delta t}}{t}} ight)\,\, = \,\, \pm \,\,\left( {\frac{{0.2}}{{13.8}}\,\, + \;\,\frac{{0.3}}{{4.0}}} ight)$

$ = \pm \,\,\left( {\frac{{0.8\,\, + \;\,4.14}}{{13.8\,\, 	imes \,\,4.0}}} ight)\,\, = \,\, \pm \,\,\frac{{4.49}}{{13.8\,\, 	imes \,\,4.0}}\,\, = \,\, \pm \,\,0.0895$

$\Delta \,v  =   \pm 0.0895\,\, 	imes \,\,v\,\, = \,\, \pm \,\,0.0895\,\, 	imes \,\,3.45\,\, = \,\, \pm 0.3087\,\, = \,\, \pm \,\,0.31$

$v  =  \left( {3.5\,\, \pm \,\,0.31} ight)\,m{s^{ - 1}}\,$

$\frac{{\Delta v}}{v}\,\, 	imes \,\,100\,\, = \,\, \pm \,\,0.0895\,\, 	imes \,\,100\,\, = \,\, \pm \,\,8.95\,\% \,\, = \,\, \pm \,\,9\% $

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

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