1. Tardigrade
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  3. Physics
  4. A boy runs towards a man standing 150 m away. Speed of boy is 5 m s-1 and that of man is 10 ms-1 The time and distance of their meeting from the boy's position are

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Q. A boy runs towards a man standing $150\, m$ away. Speed of boy is $5 \, m s^{-1}$ and that of man is $10\, ms^{-1}$ The time and distance of their meeting from the boy's position are

Motion in a Straight Line

Solution:

The relative speed = 5 + 10 = 15 m $s^{-1}$ But relative speed = $\frac{Distance}{time} $ i.e. Time = $\frac{Distance}{relative \, speed} = \frac{150}{15} = 10 s$ Distance from boy's position = velocity of boy x time = 5 $\times$ 10 = 50m