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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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प्रश्न

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

योग
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उत्तर

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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अध्याय 7: Applications of Differential Calculus - Exercise 7.1 [पृष्ठ ८]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 7 Applications of Differential Calculus
Exercise 7.1 | Q 3. (iii) | पृष्ठ ८

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