Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12   1. Chapter 6 Class 12 Application of Derivatives
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Example 42 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 𝑥=𝑡^2 (2−𝑡/3). Find the time taken by it to reach Q and also find distance b/w P & Q Given distance 𝑥 = t2 (2−𝑡/3) Let v be the velocity of the car at t second 𝑣 = change in distance w.r.t t i.e. 𝑣 = 𝑑𝑥/𝑑𝑡 𝑣 = 𝑑(𝑡^2 (2 − 𝑡/3))/𝑑𝑡 𝑣 = 𝑑(2𝑡^2− 𝑡^3/3)/𝑑𝑡 𝑣 = 4t – t2 Putting v = 0 ⇒ 4t – t2 = 0 ⇒ t(4−𝑡)=0 So, t = 0 & t = 4 Hence, the car is not moving at t = 0 & t = 4 second Now, car is not moving (i.e. v = 0) at point P as well as at point Q Thus the car will reach the point Q after 4 seconds ⇒ Distance PQ = Distance travelled in 4 seconds Finding x at t = 4 𝑥 = t2 (2−𝑡/3) 𝑥 = (4)^2 (2−4/3) = 16 ((6 − 4)/3) = 16 (2/3) = 32/3 𝑚. Hence, Distance PQ = 𝟑𝟐/𝟑 𝒎.

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