Example 30 - Chapter 6 Class 12 Application of Derivatives

Last updated at July 26, 2026 by

Ā  Example 30 - A car starts from a point P at time t = 0 seconds - Examples

part 2 - Example 30 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Example 30 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in metres, covered by it, in t seconds is given by š‘„=š‘”^2 (2āˆ’š‘”/3). Find the time taken by it to reach Q and also find distance b/w P & Q Given Distance š‘„ = t2 (2āˆ’š‘”/3) At points P and Q, the Velocity of the car is 0 Let š‘£ be the velocity of the car š‘£ = Change in Distance w.r.t ttime š’— = š’…š’™/š’…š’• Finding š’— š‘£ = š‘‘(š‘”^2 (2 āˆ’ š‘”/3))/š‘‘š‘” š‘£ = š‘‘(2š‘”^2āˆ’ š‘”^3/3)/š‘‘š‘” š‘£ = 4t ā€“ t2 Putting š’— = 0 4t ā€“ t2 = 0 t(4āˆ’š‘”)=0 So, t = 0 & t = 4 Thus, it takes 4 seconds to reach from point P to Q Also, Distance PQ = Distance travelled in 4 seconds Finding x at t = 4 š‘„ = t2 (2āˆ’š‘”/3) š‘„ = (4)^2 (2āˆ’4/3) = 16 ((6 āˆ’ 4)/3) = 16 (2/3) = 32/3 š‘š. Hence, Distance PQ = šŸ‘šŸ/šŸ‘ š’Ž.

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