A particle is projected from ground as shown. It is at point A at time $t_{1}$ , at point B at time $t_{2}$ , at point C at time $t_{3}$ & at time $t_{4}$ it is at D then

_{B} -_{A} =_{D} -_{C}

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\frac{_{A}}{t_{1}} = \frac{_{B}}{t_{2}} = \frac{_{C}}{t_{3}} = \frac{_{D}}{t_{4}}

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\frac{_{B} -_{A}}{t_{2} - t_{1}} = \frac{_{D} -_{C}}{t_{4} - t_{3}}

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\frac{_{A} +_{B}}{2} = \frac{_{C} +_{D}}{2}

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Solution

The correct option is C $\frac{_{B} -_{A}}{t_{2} - t_{1}} = \frac{_{D} -_{C}}{t_{4} - t_{3}}$
During motion $\to a$ is constant hence $\frac{- - \to Δ V}{Δ t}$ is same for all intervals.

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A particle is projected from ground as shown. It is at point A at time t1, at point B at time t2, at point C at time t3 & at time t4 it is at D then

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