It is given that xvdtdv=9t2−t3=dtdx=18t−3t2=a=18−6t for maximum speed dtdv=0 and dt2d2v negative So 18−6t=0 ⇒t=3s at t=3s,x=9(3)2−(3)3 =81−27=54 m
The position x of a particle with respect to time t along x-axis is given by x=9t2 −t3 where x is in metres and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction?
Held on 30 Apr 2007 · Verified 9 Jul 2026.
54 m
81 m
24 m
32 m.
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