The position of a particle is given by r vector=2ti+3t2j+4k find the velocity and acceleration of particle at t=4s
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This is a mixed conceptual question from derivatives of vectors and understanding of vectors.
The differentiation of vectors is same as that of normal equations.
Hence we can find the resistance vector and differentiate it to get velocity as a function of time and then differentiate it again to get acceleration as a function of time.
So r=2t(i)+3t^2(j)+4(k)
Hence for velocity we differentiate r with respect to t
:
v=dr/dt=2dt/dt (i)+3dt^2/dt (j)+d4/dt (k)+0(k). at t=4s
=> v=dr/dt=2(i)+6t(j) at t=4 is _____(1)
v=2(i)+24(j)+0(k)
Now for acceleration we can differentiate eq(1) once again to obtain that
a=dv/dt= d2 (i)+6dt/dt (j) +d0dt (k)
=> a=0(i)+6(j)+0(k)
Hope it helps :)
The differentiation of vectors is same as that of normal equations.
Hence we can find the resistance vector and differentiate it to get velocity as a function of time and then differentiate it again to get acceleration as a function of time.
So r=2t(i)+3t^2(j)+4(k)
Hence for velocity we differentiate r with respect to t
:
v=dr/dt=2dt/dt (i)+3dt^2/dt (j)+d4/dt (k)+0(k). at t=4s
=> v=dr/dt=2(i)+6t(j) at t=4 is _____(1)
v=2(i)+24(j)+0(k)
Now for acceleration we can differentiate eq(1) once again to obtain that
a=dv/dt= d2 (i)+6dt/dt (j) +d0dt (k)
=> a=0(i)+6(j)+0(k)
Hope it helps :)
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