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Question
The distance travelled by an object in time (100 ± 1)s is (5.2 ± 0.1) m. What are the speed and its maximum relative error?
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Solution
Given: Distance (D ± ΔD) = (5.2 ± 0.1) m, time (t ± Δt) = (100 ± 1) s.
To find: Speed (v), the maximum relative error `((triangle"v")/"v")`
Step 1- Calculating Speed
Given - D = 5.2 m, t = 100 s
v = `"D"/"t"`
`= (5.2 " m")/(100 " s")`
= 0.052 m s-1
Step 2 - Calculating relative error
Given - ΔD = 0.1 m; Δt = 1 s
`(triangle"v")/"v" = +- ((triangle"D")/"D" + (triangle"t")/"t")`
`(triangle"v")/"v" = +- ((0.1 " m")/(5.2 " m") + (1 " s")/(100 " s"))`
`= +- (1/52 + 1/100)`
`= +- (100 + 52)/5200`
`= +- 152/5200 " i.e." +- 19/650`
= ± 0.0292 m s-1
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