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A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. At what times the particle changes direction?

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प्रश्न

A particle moves along a line according to the law s(t) = 2t3 – 9t2 + 12t – 4, where t ≥ 0. At what times the particle changes direction?

योग
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उत्तर

s = f(t) = 2t3 – 9t2 + 12t – 4

V = f'(t) = 6t2 – 18t + 12

V = 0 ⇒ 6(t2 – 3t – 2) = 0

(t – 1)(t – 2) = 0

t = 1, 2

When t < 1, (say t = 0.5)

V = 6(0.25 – 1.5 + 2) = +ve

When 1 < t < 2, (say t = 1.5)

V = 6(2.25 – 4.5 + 2) = – ve

When t > 2, (say t = 3)

V = 6(9 – 6 + 2) = +ve

So the particle changes its direction when t lies between 1 and 2 secs.

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Meaning of Derivatives
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Applications of Differential Calculus - Exercise 7.1 [पृष्ठ ८]

APPEARS IN

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

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