Application of Derivatives

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 Find the time taken by it to reach Q and also find distance between P and Q. - Wired Faculty

Question

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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
$straight x space equals straight t squared open parentheses 2 minus straight t over 3 close parentheses.$
Find the time taken by it to reach Q and also find distance between P and Q.

Solution

space space space space space space space space space space space space space space space space space space space space straight x space equals space straight t squared open parentheses 2 minus straight t over 3 close parentheses space equals space 2 space straight t squared minus straight t squared over 3 therefore space space space space space space space space dx over dt space equals space 4 straight t space minus space straight t squared space equals space straight t left parenthesis 4 minus straight t right parenthesis therefore space space space space space space space space space space space space straight v space equals space straight t left parenthesis 4 minus straight t right parenthesis Now comma space space space space straight v space equals space 0 space space space space space space space space space space rightwards double arrow space space space space space space space straight t left parenthesis 4 minus straight t right parenthesis space equals space 0 space space space space space space space space space space space space space rightwards double arrow space space space space straight t space equals space 0 comma space space 4 Now space space space space space straight t space equals space 0 space is space the space time space at space starting space point space straight P space therefore space space space space space space space straight t space equals space 4 space seconds space is space the space time space to space recover space distance space PQ. thin space PQ space equals space distance space covered space in space 4 space seconds space space space space space space space space space space space space space space space space space space space space equals space open parentheses 4 close parentheses squared space open square brackets 2 minus 4 over 3 close square brackets space equals space 16 space open parentheses 2 over 3 close parentheses space equals space 32 over 3 therefore space space space space space PQ space equals space 32 over 3 space metres.

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Application Of Derivatives

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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