Question

two particles P and Q are initially 40m apart P behind Q. particle P starts moving with a uniform velocity 10m/s towards Q. particle Q starting from rest has an acceleration 2m/s2 in the direction of velocity of P. then the min distance between P and Q will be

Answer 1

$$\begin{gathered} Thedis\tan cecoveredbyPintimet=S_P=v\times t=10\times t \\ Thedis\tan cecoveredbyQintimet=S_Q=\frac{1}{2}\times 2\times t^2=t^2 \\ AftertimettheseperationbetweenPandQ=D=40+t^2-10t \\ FOrdis\tan cebetweenthetwotobeminimum\frac{dD}{dt}=0 \\ Therefore\frac{dD}{dt}=0+2t-10=0 \\ ort=5s \\ Thustheminimumdis\tan cebetweenthetwo=D=40+t^2-10t=40+25-50=15m. \end{gathered}$$