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Question

Physics Question on Units and measurement

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

A

(3.5±0.6)ms1\left(3.5 \pm 0.6\right)\,ms^{-1}

B

(3.5±0.3)ms1\left(3.5 \pm 0.3\right)\,ms^{-1}

C

(6.1±0.6)ms1\left(6.1 \pm 0.6\right)\,ms^{-1}

D

(6.1±0.3)ms1\left(6.1 \pm 0.3\right)\,ms^{-1}

Answer

(3.5±0.3)ms1\left(3.5 \pm 0.3\right)\,ms^{-1}

Explanation

Solution

Here, s=(13.8±0.2)ms=\left(13.8 \pm0.2\right)\,m, t=(4.0±0.3)st=\left(4.0 \pm 0.3\right)\,s \therefore veliocity, v=st=13.84.0=3.45ms1v=\frac{s}{t}=\frac{13.8}{4.0}=3.45\,ms^{-1} (Rounded off to first place of decimal) Δvv=Δss+Δtt=0.213.8+0.34.0\therefore \frac{\Delta v}{v}=\frac{\Delta s}{s}+\frac{\Delta t}{t}=\frac{0.2}{13.8}+\frac{0.3}{4.0} =0.8+4.1413.8×4.0=4.9413.8×4.0=\frac{0.8+4.14}{13.8\times4.0}=\frac{4.94}{13.8 \times 4.0} =0.0865=0.0865 or Δv=v×0.0865=3.45×0.0865\Delta v=v\times 0.0865=3.45 \times 0.0865 =0.3087=0.3087. \therefore velocity =(3.5±0.3)ms1=\left(3.5 \pm 0.3\right)\,ms^{-1}.