A particle position in the xy plane varies as x(t) =3t3+4t2,y(t)=16t4+2. find the acceleration vector of the particles after 1 second of start
Answers
Answered by
1
Vx= 9t^2+8t
Vy=64t^3+0
By differentiate
Again differentiate we will get acceleration so
Ax= 18t+8
Ay=192t^2 now put t=1
And take its magnitude
By (Ax^2+Ay^2)^1/2
Vy=64t^3+0
By differentiate
Again differentiate we will get acceleration so
Ax= 18t+8
Ay=192t^2 now put t=1
And take its magnitude
By (Ax^2+Ay^2)^1/2
SwathiPriya681:
thnk u
Similar questions