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Distance vs Time Graph

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Question

From the displacement-time graph of a cyclist given in the figure, find the average velocity in the first 4 s.

0.5 m s^{- 1}

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5 m s^{- 1}

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2.5 m s^{- 1}

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0 m s^{- 1}

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Solution

The correct option is C $2.5 m s^{- 1}$
Total displacement in first 4 s = 10 m.
Therefore, average velocity in first 4 s, $v = \frac{D i s p l a c e m e n t}{T i m e} = \frac{10 m}{4 s} = 2.5 m / s$

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Similar questions

From the displacement-time graph of a cyclist, given in the figure, find:

(i) the average velocity in the first 4 s,

(ii) the displacement from the initial position at the end of 10 s,

(iii) the time after which he reaches the starting point.

t = 2 s

t = 5 s

m

s

- 2.5 m s^{- 1}

10 s

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