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

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=100t=10050=2 sNow when y=0 then ,0=100-4t2100=4t2t2=25t=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

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