The graph given below is the distance-time graph of an object.
$(i)$ Find the speed of the object during the first four seconds of its journey.
(ii) How long was it stationary?
(iii) Does it represent a real-life situation? Justify your answer.
$(i)$ Find the speed of the object during the first four seconds of its journey.
(ii) How long was it stationary?
(iii) Does it represent a real-life situation? Justify your answer.
(N/A) $(i)$ Speed is the slope of the distance-time graph. For the first $4 \ s$,the distance covered is $75 \ m$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{75 \ m}{4 \ s} = 18.75 \ m \ s^{-1}$.
(ii) The object is stationary when the distance remains constant over time. This corresponds to the horizontal line segment $PQ$. The object is stationary from $t = 4 \ s$ to $t = 14 \ s$.
Duration $= 14 \ s - 4 \ s = 10 \ s$.
(iii) No,this graph does not represent a real-life situation. In a real-life scenario,the distance covered by an object cannot decrease with time. The segment $RQ$ shows the object moving from a distance of $0 \ m$ to $75 \ m$ between $10 \ s$ and $14 \ s$,but the graph implies the object was at $75 \ m$ at $t=4 \ s$ and then suddenly appears at $0 \ m$ at $t=10 \ s$,which is physically impossible.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{75 \ m}{4 \ s} = 18.75 \ m \ s^{-1}$.
(ii) The object is stationary when the distance remains constant over time. This corresponds to the horizontal line segment $PQ$. The object is stationary from $t = 4 \ s$ to $t = 14 \ s$.
Duration $= 14 \ s - 4 \ s = 10 \ s$.
(iii) No,this graph does not represent a real-life situation. In a real-life scenario,the distance covered by an object cannot decrease with time. The segment $RQ$ shows the object moving from a distance of $0 \ m$ to $75 \ m$ between $10 \ s$ and $14 \ s$,but the graph implies the object was at $75 \ m$ at $t=4 \ s$ and then suddenly appears at $0 \ m$ at $t=10 \ s$,which is physically impossible.
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