First slide

Instantaneous acceleration

Question

The displacement of a particle moving along the x-axis is given by equation x = 2 $t^{3}$ – 21 $t^{2}$ + 60t + 6. The possible acceleration of the particle when its velocity is zero is

Moderate

A

–18m/ $s^{2}$
B

30 m/ $s^{2}$
C

9m/ $s^{2}$
D

–9m/ $s^{2}$

Solution

$\large x = 2{t^3} - 21{t^2} + 60t + 6$
$\large \frac{{dx}}{{dt}} = V = 6{t^2} - 42t + 60$
$\large 0 = 6{t^2} - 42t + 60;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} t = 2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} 5{\kern 1pt} {\kern 1pt} \sec$
$\large \frac{{dv}}{{dt}} = a = 12t - 42$
$\large {\text{t = 2sec; a = }} - 18{\kern 1pt} {\kern 1pt} m/{\sec ^2}$

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