Civil Engineering - UPSC Civil Service Exam Questions - Discussion

V= (6+2xy+t2) i

Acceleration at x axis = (udu/dx) + (du/dt)------>(a)
(6+ 2xy+t2)= u ------------->(1)
du/dx= 2y ------------------>(2)
du/dt= 2t ------------------->(3)

Put 1,2,3 in equation (a).
(6+2xy+t2)*2y+(2t)---------->(b)
Points given ( 3,1,2) at time =2Subsitute points and time in equation (b).
Ans 36 units.

As u know,

V=ui+vj for 2-D (because velocity in y and z-direction is not given)
Therefore , x component is given that is u =6+2xy+t^2.
Then acceleration along x-direction ax = du/dt =du/dt+udu/dx +vdu/dy +wdu/dz .
Here last two-part (vdu/dy and wdu/dz) would be zero.

Then, 2t+(6+2xy+t^2)*2y then t = 2sec and(3,1,2) finally put this value in above then we get ax=36 unit.

V= ui + vj + wz.

But here only x component is given that is u.
Acceleration in x direction= du/dt= udu/dx+vdu/dt+wdu/dz+du/dt.
So, Ax= (6+2xy+t^2)*2y+0+0+2t = 36 units.

Please explain the answer in detail.