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Example 42 - A car starts from a point P at time t = 0 seconds

Example 42 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 42 - Chapter 6 Class 12 Application of Derivatives - Part 3

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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 π‘₯=𝑑^2 (2βˆ’π‘‘/3). Find the time taken by it to reach Q and also find distance b/w P & Q Given Distance π‘₯ = t2 (2βˆ’π‘‘/3) At points P and Q, the Velocity of the car is 0 Let 𝑣 be the velocity of the car 𝑣 = Change in Distance w.r.t ttime 𝒗 = 𝒅𝒙/𝒅𝒕 Finding 𝒗 𝑣 = 𝑑(𝑑^2 (2 βˆ’ 𝑑/3))/𝑑𝑑 𝑣 = 𝑑(2𝑑^2βˆ’ 𝑑^3/3)/𝑑𝑑 𝑣 = 4t – t2 Putting 𝒗 = 0 4t – t2 = 0 t(4βˆ’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 (2βˆ’π‘‘/3) π‘₯ = (4)^2 (2βˆ’4/3) = 16 ((6 βˆ’ 4)/3) = 16 (2/3) = 32/3 π‘š. Hence, Distance PQ = πŸ‘πŸ/πŸ‘ π’Ž.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.