The displacement of a particle moving along the x-axis is given by equation x=2t³–21t²+60t+6. The possible acceleration of the particle when its velocity is zero is

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30 m/s²

–18m/s²

9m/s²

–9m/s²

answer is A.

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Detailed Solution

x=2t³–21t²+60t+6 ;
$\frac{dx}{dt} = v = 6 t^{2} - 42 t + 60$
0 = 6t² - 42t + 60 ; t = 2 or 5 sec
$\frac{dv}{dt} = a = 12 t - 42$
t = 2sec; a = – 18m/sec²

at t=5 sec, a= +18m/s².

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