Finding rate of change

Chapter 6 Class 12 Application of Derivatives (Term 1)
Concept wise

### Transcript

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 π₯=π‘^2 (2βπ‘/3). Find the time taken by it to reach Q and also find distance b/w P & Q Given Distance π₯ = t2 (2βπ‘/3) At points P and Q, the Velocity of the car is 0 Let π£ be the velocity of the car π£ = Change in Distance w.r.t ttime π = ππ/ππ Finding π π£ = π(π‘^2 (2 β π‘/3))/ππ‘ π£ = π(2π‘^2β π‘^3/3)/ππ‘ π£ = 4t β t2 Putting π = 0 4t β t2 = 0 t(4βπ‘)=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 π₯ = t2 (2βπ‘/3) π₯ = (4)^2 (2β4/3) = 16 ((6 β 4)/3) = 16 (2/3) = 32/3 π. Hence, Distance PQ = ππ/π π.

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#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.