Question

Velocity of a particle in x-y plane at any time t is v=(2t hati+3t^2 hatj) m//s At t=0, particle starts from the co-ordinates (2m,4m). Find (a) acceleration of the particle at t=1s. (b) position vector and co-ordinates of the particle at t=2s.

Answer 1

Explanation:

v at 1 sec = 3i + j

x = 7/3 i + 1/4 j

Explanation:

acceleration of a particle in x-y plane varies with time as a(2 ti+3t2j)m/s2 at time t=o, velocity of the particle is 2m/s.along positive x-direction and particle starts from origin find velocity and coordinates of particle at t=1s

a = 2t i  + 3t²j

a = dv/dt

=> v = adt

v = ∫adt

=> v  =  t²i + t³j + c

at t = 0 v = 2i m/s

=> 2i = 0 + 0 + c

=> c = 2i

v = t²i + t³j + 2i

at t = 1

V = i + j + 2i

v at 1 sec = 3i + j

v = t²i + t³j + 2i

x = ∫vdt

x = t³/3i + t⁴/4j  + 2ti  + c

at t = 0 x = 0

0 = 0 + 0 + 0 + c

=> c = 0

x = t³/3i + t⁴/4j  + 2ti

at t = 1

x = 1/3i + 1/4j + 2t

x = 7/3 i + 1/4 j

Answer 2

Answer:

x = 7/3i, 4/3 j.

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