At zero acceleration, a particle moves along a straight line. What is the velocity of a particle at the position of x = t4−20t2+42?
A
-48.68 m/sec
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B
+48.68 m/sec
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C
-42 m/sec
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D
+ 42 m/sec
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Solution
The correct option is A -48.68 m/sec x=t4−20t2+42
Velocity =dxdt=4t3−40t
Acceleration =d2xdt2=12t2−40
where acceleration is zero. 12t2=40 t=√4012sec
Velocity =4(4012)3/2−40(4012)1/2 =(24−73.03)m/sec =−48.63m/sec