Physics

From the displacement-time graph of a cyclist, given in 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.

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(i) As we know,

$\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}}$

Substituting the values from the graph, for the first 4 s, we get,

$\text {Avg. velocity} = \dfrac {10 \text {m}}{4 \text {s}} \\[0.5em] \Rightarrow \text {Avg. velocity} = 2.5\text {ms} ^ {-1} \\[0.5em]$

Hence, average velocity of the car is 2.5 ms^-1.

(ii) From the graph, we get,

Displacement = final position - initial position

Substituting the values from the graph, for the first 10 s, we get,

Displacement = -10 - 0 = -10 m

Hence, displacement from the initial position at the end of 10 s = - 10 m

(iii) The cyclist would reach the start point two times, one at the 7 s and the other at 13 s.

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