A particle starts from origin at t=0 witha vel of 8j m/s and moves in x-y plane with const.acclrtn of (4i+2j)m/s².At the instant the particle's x-coordinate is 29m,what are

1)its y-coordinate?

2)Its speed?

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Dear Chilukuri

for motion in x direction

Ux=0

a= 4 m/sec2

Sx= 29 m

Sx=Ux t+1/2 a t²

29 =0+1/2 * 4 * t²

t² =29/2  ........................1

Vx= Ux +at

  Vx= 4t ......................2

Motion in y direction

Sy = Uyt+1/2 a t²

Sy =8 t +1/2 * 2* t²

    =8t +t²               Answer

put value of t from equation 1 and find Sy

Vy =Uy + at

     =8  +2t   put value of t from equation 1 and find Vy       

now speed at this point =( Vx² +Vy²)^1/2                 Answer

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