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Motion in A Plane

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12

The position x of a particle with respect to time t along x- axis is given by x = 9t2 -t3 where x is in meter and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction? 

  • 32 m

  • 54 m

  • 81 m 

  • 24 m


B.

54 m

At the instant when speed is maximum, its acceleration is zero.

Given, the position x of particle with respect to time t along x- axis

x = 9t2-t3 ... (i)
differentiating Eq. (i) with respect to time, we get speed, ie,
straight v space equals space dx over dt space equals space straight d over dt left parenthesis 9 straight t squared minus straight t cubed space right parenthesis
straight v space equals space 18 space straight t space minus space 3 straight t squared space... space left parenthesis ii right parenthesis

Again differentiating Eq. (ii) with respect to time, we get acceleration, ie,

straight a space equals space dv over dt space equals straight d over dt left parenthesis 9 straight t squared minus straight t cubed right parenthesis
straight a space equals space 18 minus 6 straight t
Now, when speed of particle is maximum, its acceleration is zero, ie,

a= 0
18-6t = 0 or t = 3s
Putting in eq (i) We, obtain the position of a particle at that time. 

x = 9 (3)2 - (3)3 = 9 (9) -27 
= 81-27 = 54 m

At the instant when speed is maximum, its acceleration is zero.

Given, the position x of particle with respect to time t along x- axis

x = 9t2-t3 ... (i)
differentiating Eq. (i) with respect to time, we get speed, ie,
straight v space equals space dx over dt space equals space straight d over dt left parenthesis 9 straight t squared minus straight t cubed space right parenthesis
straight v space equals space 18 space straight t space minus space 3 straight t squared space... space left parenthesis ii right parenthesis

Again differentiating Eq. (ii) with respect to time, we get acceleration, ie,

straight a space equals space dv over dt space equals straight d over dt left parenthesis 9 straight t squared minus straight t cubed right parenthesis
straight a space equals space 18 minus 6 straight t
Now, when speed of particle is maximum, its acceleration is zero, ie,

a= 0
18-6t = 0 or t = 3s
Putting in eq (i) We, obtain the position of a particle at that time. 

x = 9 (3)2 - (3)3 = 9 (9) -27 
= 81-27 = 54 m

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