The position $x$ of a particle varies with time $(t)$ as $x=P t^{3}-Q t^{4}$. The acceleration will

Question

The position $x$ of a particle varies with time $(t)$ as $x=P t^{3}-Q t^{4}$. The acceleration will be zero at time $t$ equal to

Professor PC Thomas Classes · Answer

(a) : Distance $(x)=P t^{3}-Q t^{4}$ \ Therefore velocity $(v)=\frac{d x}{d t}=3 P t^{2}-4 Q t^{3}$ \ Acceleration $=\frac{d v}{d t}=6 P t-12 Q t^{2}$ \ For acceleration to be zero: $6 P t-12 Q t^{2}=0 \implies 6 P t = 12 Q t^{2} \implies t=\frac{6 P}{12 Q}=\frac{P}{2 Q}$