A particle moves in xy plane such that v x = 50 - 16 t and y = 100 - 4 t 2 where v x is in m s 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 Share with your friends Share 4 Pintu B. answered this Dear Student , Here in this case the given data is , vx=50-16tand y=100-4t2Hence the first and second equation can be written as ,v=u-atS=ut-12at2Now comparing both sides we get ,u=50 m/s and ut=100⇒t=10050=2 sNow when y=0 then ,0=100-4t2⇒100=4t2⇒t2=25⇒t=5 sSo hence when y=0 then velocity ,vx=50-80=-30 m/s .negative sign shows that velocity is decreasing with increase in time So the acceleration of the particle is ,a=v+ut=-30+502=202=10 m/s2 Regards 8 View Full Answer