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Tamil Nadu Board of Secondary EducationHSC Science Class 12

A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. Find the particle’s acceleration each time the velocity is zero

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Question

A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. Find the particle’s acceleration each time the velocity is zero

Sum
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Solution

s(t) = 2t3 – 9t2 + 12t – 4

Velocity v = `"ds"/"dt"`

= 6t2 – 18t + 12

v = 0

⇒ 6(t2 – 3t + 2) = 0

⇒ t = 1, 2

Acceleration = `("d"^2"s")/("dt"^2)`

= 12t – 18

At t = 1, Acceleration = 12(1) – 18 = – 6 m/sec2

At t = 2, Acceleration = 12 (2) – 18 = 6 m/sec2

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Chapter 7: Applications of Differential Calculus - Exercise 7.1 [Page 8]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.1 | Q 3. (iii) | Page 8

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