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Acceleration

A particle st...

Question

A particle starts moving along the x-axis from $t = 0$ , its position varying with time as $x = 2 t^{3} - 3 t^{2} + 1$ ,
a. At what time instants is its velocity zero?
b. What is the velocity when it passes through the origin?

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Solution

$x = 2 t^{3} - 3 t^{2} + 1 \Rightarrow v = \frac{d x}{d t} = 6 t^{2} - 6 t$
a. Put $v = 0 \Rightarrow 0 = 6 t^{2} - 6 t \Rightarrow t = 1 s, t = 0 s$
b. $x = 0 \Rightarrow 0 = 2 t^{3} - 3 t^{2} + 1 \Rightarrow (t - 1)^{2} (2 t + 1) = 0$
$\Rightarrow t = 1 s, v_{t = 1 s} = 6 \times 1^{2} - 6 \times 1 = 0 m s^{- 1}$

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