Question

A particle moves in xy plane such that $v_x=50-16t$ and $y=100-4t^2$ where $v_x$ is in $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ and y is in m. It is also known that x=0 when t-0. Determine (i) Acceleration of particle (ii) Velocity of particle when y=0

Answer 1

Dear Student ,
Here in this case the given data is ,

$$\begin{gathered} v_x=50-16t \\ andy=100-4t^2 \\ Hencethefirstandsecondequationcanbewrittenas, \\ v=u-at \\ S=ut-\frac{1}{2}a{t}^2 \\ Nowcomparingbothsidesweget, \\ u=50m/sand \\ ut=100 \\ \Rightarrow t=\frac{100}{50}=2s \\ Nowwheny=0then, \\ 0=100-4t^2 \\ \Rightarrow 100=4t^2 \\ \Rightarrow t^2=25 \\ \Rightarrow t=5s \\ Sohencewheny=0thenvelocity,v_x=50-80=-30m/s. \\ \left[negativesignshowsthatvelocityisdecrea\sin gwithincreaseintime\right] \\ Sotheaccelerationoftheparticleis, \\ a=\frac{v+u}{t}=\frac{-30+50}{2}=\frac{20}{2}=10m/s^2 \end{gathered}$$

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