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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? - Physics

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प्रश्न

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?

संख्यात्मक
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उत्तर

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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अध्याय 1: Units and Measurements - Exercises [पृष्ठ १४]

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बालभारती Physics [English] Standard 11 Maharashtra State Board
अध्याय 1 Units and Measurements
Exercises | Q 3. ii) | पृष्ठ १४

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