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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?
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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