# Example 42

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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 𝑥=𝑡22−𝑡3. Find the time taken by it to reach Q and also find distance between P and Q Given distance 𝑥 = t2 2−𝑡3 Let v be the velocity of the car at t second 𝑣 = change in distance w.r.t t i.e. 𝑣 = 𝑑𝑥𝑑𝑡 𝑣 = 𝑑𝑡22 − 𝑡3𝑑𝑡 𝑣 = 𝑑2𝑡2− 𝑡33𝑑𝑡 𝑣 = 4t – t2 Putting v = 0 ⇒ 4t – t2 = 0 ⇒ t4−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 i.e. 𝑥 = t2 2−𝑡3 at t = 4 𝑥𝑡=4 = 422−43 = 16 6 − 43 = 16 23 = 323 𝑚. Hence, Distance PQ = 𝟑𝟐𝟑 𝒎.

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Chapter 6 Class 12 Application of Derivatives

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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 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.