Chapter 4
Describing Motion Around Us
CBSE Class 9
Science Solutions
Educart Science class Class 9 NCERT Exemplar cover
Question:

A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The given graph depicts her velocity versus time. Estimate the distance she ran based on the graph.

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Answer: Verified

From the graph:

0 to 1 h: velocity increases from 7 to 7.5 km h1^{-1}

1 to 3 h: velocity remains 7.5 km h1^{-1}

3 to 6 h: velocity decreases from 7.5 to 6.5 km h1^{-1}

So, From 0 h to 1.6 h,

Distance=7×0.6=4.2 km\text{Distance} = 7 \times 0.6 = 4.2 \text{ km}

From 0.6 h to 1.6 h,

Distance=(7+7.5)2×1=7.25 km\text{Distance} = \frac{(7+7.5)}{2} \times 1 = 7.25 \text{ km}

From 1.6 h to 3.0 h,

Distance=7.5×1.4=10.5 km\text{Distance} = 7.5 \times 1.4 = 10.5 \text{ km}

From 3.0 h to 4.6 h,

Distance=(7.5+7)2×1.6=11.6 km\text{Distance} = \frac{(7.5+7)}{2} \times 1.6 = 11.6 \text{ km}

From 4.6 h to 5.6 h,

Distance=(7+6.5)2×1=13.5 km\text{Distance} = \frac{(7+6.5)}{2} \times 1 = 13.5 \text{ km}

From 5.6 h to 6.6 h,

Distance=6.5×1=6.5 km\text{Distance} = 6.5 \times 1 = 6.5 \text{ km}

Therefore,

Total distance=4.2+7.25+10.5+11.6+13.5+6.5=53.55 km\begin{aligned} \text{Total distance} &= 4.2 + 7.25 + 10.5 + 11.6 + 13.5 \\ &\quad + 6.5 \\ &= 53.55 \text{ km} \end{aligned}

Distance travelled ≈ 53.55 km

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