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Application Of Derivatives

Question
CBSEENMA12035507
Wired Faculty App

The space s described in time t by a particle moving in a straight line is given by s = r5 – 40r + 30r2 + 80t – 250. Find the minimum value of its acceleration.

 

Solution

Here,    straight s space equals space straight t to the power of 5 space minus space 40 space straight t cubed space minus space 30 space straight t squared space plus space 80 space straight t space minus space 250
therefore space space space ds over dt space equals space 5 straight t to the power of 4 minus 120 space straight t squared plus 60 straight t plus 80 space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space 20 straight t cubed minus 240 space straight t space plus 60 therefore space space space acceleration space space straight f space equals space 20 space straight t cubed space minus 240 space straight t space plus space 60 space space space space space df over dt space equals 60 space straight t squared minus 240 space space space space space space df over dt space equals space 0 space space space space space space space space space rightwards double arrow space space space 60 straight t squared minus 240 space equals space 0 space space space space space rightwards double arrow space space space space straight t squared minus 4 space equals space 0 space space space space space rightwards double arrow space space space straight t squared space equals space 4 space space space space rightwards double arrow space space space space straight t space space equals negative 2 comma space 2 But space space space straight t not equal to space minus 2 comma space space space space space as space straight t greater than 0 space space therefore space space space space space space straight t space equals space 2 space space space space fraction numerator straight d squared straight f over denominator dt squared end fraction space equals space 120 space straight t At space straight t space space equals 2 comma space space space space space fraction numerator straight d squared straight f over denominator dt squared end fraction space equals space 240 space greater than 0 therefore space space space space space space straight f space is space minimum space at space straight t space equals space 2 therefore space space space space space space minimum space straight f space equals space 20 left parenthesis 2 right parenthesis cubed space minus 240 left parenthesis 2 right parenthesis space plus space 60 space equals space space 160 minus 480 plus 60 equals negative 260 space space space space space space space space space space

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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