Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A particle mover along the parabolic path x=y^2 + 2y + 2 in such a way that the y-component of velocity vector remains 5 m/s during the motion.The magnitude of acceleration of the particle is-
Hi, A = root{(Ax)^2 + (Ay)^2} A= total acceleration Ax = acceleration along x axis = d^2x/dt^2 Ay = acceleration along y axis = d^2y/dt^2 here vel. in y directn =dy/dt = 5 m/s (constant) => Ay = 0 now differentiate the given eqn => Vx = 2y*dy/dt +2* dy/dt = 2y*5 + 2 *5 Vx = 10y + 10 again diff. Ax = 10 *dy/dt = 10* 5 = 50 m/s^2 A = root{(Ax)^2 + (Ay)^2} => A = Ax = 50 (as Ay=0 )
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -