Position of particle is given by $x = t^{3} - 4 t^{2} + 5 t + 9.$ What would be the distance traveled by particle from instant $t = 0$ to instant when particle changes its direction of velocity for last time.

1.85 m

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2.15 m

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2 m

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2.85 m

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Solution

The correct option is B $2.15 m$
$\begin{matrix} x = t^{3} - 4 t^{2} + 5 t + 9 a t t = 0 x = 9 v = 3 t^{2} - 8 t + 5 = 0 t = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm 2}{6} = \frac{10}{6}, 1 = \frac{5}{3}, 1 a t t = 1 x = 1 - 4 + 5 + 9 = 11 a t t = 3 x = \frac{125}{27} - \frac{4 \times 25}{9} + \frac{25}{3} + 9 = 10.85 ∴ T o t a l d i s t = (11 - 9) + (11 - 10.85) = 2.15 m \end{matrix}$
Hence,
option $(B)$ is correct answer.

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Similar questions

x = t^{3} - 12 t^{2} + 6 t + 8

x = 2 + 8 t - 4 t^{2}

A particle travels along the X-axis. Its velocity time graph is shown above.

At what time instant did the particle change its direction of motion?

t = 0

T

A

t = 0

t = 5 T / 4

0 \leq t \leq 20

y (t) = 5 t - \frac{t^{2}}{3}

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