The distance covered by a body in time 5.0±0.6S is 40.0±0.4m . Velocity of the body in ms−1 is

velocity v=st=40.05.0=8.0 m/sΔv=Δssv+Δttv∆V=0.440+0.658=1.04∴v±Δv=8.0±1.04 m/s

The distance covered by a body in time 5.0±0.6S is 40.0±0.4m . Velocity of the body in ms−1 is

velocity v=st=40.05.0=8.0 m/sΔv=Δssv+Δttv∆V=0.440+0.658=1.04∴v±Δv=8.0±1.04 m/s

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The distance covered by a body in time $(5.0 \pm 0.6) S$ is $(40.0 \pm 0.4) m$ . Velocity of the body in ${ms}^{- 1}$ is

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$(8.0 \pm 1.04)$

$(8.0 \pm 0.6)$

$(8.0 \pm 1.0)$

$(8.0 \pm 0.4)$

answer is A.

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$v e l o c i t y v = \frac{s}{t} = \frac{40.0}{5.0} = 8.0 m / s$

$Δ v = (\frac{Δ s}{s}) v + (\frac{Δ t}{t}) v$

$∆ V = (\frac{0.4}{40} + \frac{0.6}{5}) 8 = 1.04$

$∴ v \pm Δ v = (8.0 \pm 1.04) m / s$

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The distance covered by a body in time 5.0±0.6S is 40.0±0.4m . Velocity of the body in ms−1 is

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The distance covered by a body in time $(5.0 \pm 0.6) S$ is $(40.0 \pm 0.4) m$ . Velocity of the body in ${ms}^{- 1}$ is