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Example 42 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
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 ๐ฅ=๐ก^2 (2โ๐ก/3). Find the time taken by it to reach Q and also find distance b/w P & Q Given distance ๐ฅ = t2 (2โ๐ก/3) Let v be the velocity of the car at t second ๐ฃ = change in distance w.r.t t i.e. ๐ฃ = ๐๐ฅ/๐๐ก ๐ฃ = ๐(๐ก^2 (2 โ ๐ก/3))/๐๐ก ๐ฃ = ๐(2๐ก^2โ ๐ก^3/3)/๐๐ก ๐ฃ = 4t โ t2 Putting v = 0 โ 4t โ t2 = 0 โ t(4โ๐ก)=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 โ Distance PQ = Distance travelled in 4 seconds Finding x at t = 4 ๐ฅ = t2 (2โ๐ก/3) ๐ฅ = (4)^2 (2โ4/3) = 16 ((6 โ 4)/3) = 16 (2/3) = 32/3 ๐. Hence, Distance PQ = ๐๐/๐ ๐.
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.