Skip to content
Numerical · Q13

Q.The distance travelled by an object in time (100 ± 1) s is (5.2 ± 0.1) m. What is the speed and it's relative error?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
48% · 13/27 Questions
✓ Free question

Step 1. Speed v=st=5.2 m100 s=0.052 m s−1v = \dfrac{s}{t} = \dfrac{5.2\ \text{m}}{100\ \text{s}} = 0.052\ \text{m s}^{-1}.

Step 2. By the division-error rule (Equation 1.7), the relative error in vv is the sum of the relative errors in ss and tt: Δvv=Δss+Δtt=0.15.2+1100=0.01923+0.01=0.02923\dfrac{\Delta v}{v} = \dfrac{\Delta s}{s} + \dfrac{\Delta t}{t} = \dfrac{0.1}{5.2} + \dfrac{1}{100} = 0.01923 + 0.01 = 0.02923.

Step 3. So the relative error is about 0.02920.0292, i.e. a percentage error of about 2.92%2.92\%. Note: the printed textbook answer states this relative error with a unit attached ('± 0.0292 m s⁻¹'), but relative error is by definition a dimensionless ratio (Equation 1.5) and should not carry a unit — the absolute error would be Δv=v×(Δv/v)≈0.052×0.0292≈0.0015 m s−1\Delta v = v \times (\Delta v/v) \approx 0.052 \times 0.0292 \approx 0.0015\ \text{m s}^{-1}, which is the quantity that correctly carries the m/s unit.

✓Final answer

Speed = 0.052 m s−10.052\ \text{m s}^{-1}; relative error ≈0.0292\approx 0.0292 (a dimensionless ratio, i.e. about 2.92%).

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.