Question

a particle moves along the space curve r=(t^2+t)i+(3t-2)j+(2t^3-4t^2)k (t in second ,r in m) find at time t=2 the a) velocity b) acceleration c) speed or magnitude of velocity and magnitude of acceleration

Answer 1

Answer:

V= √98

a = 2√65

Explanation:

a particle moves along the space curve r=(t^2+t)i+(3t-2)j+(2t^3-4t^2)k (t in second ,r in m) find at time t=2 the a) velocity b) acceleration c) speed or magnitude of velocity and magnitude of acceleration

r = (t² + t)i  + (3t - 2)j  + (2t³ - 4t²)k

X =  (t² + t)i  + (3t - 2)j  + (2t³ - 4t²)k

Velocity = dX/dt  = (2t + 1)i + 3j  + (6t² -8t)k

Velocity at t = 2  

= ( 2*2 + 1)i + 3j + (6*2² - 8*2)k

= 5i + 3j + 8k

Magnitude of Velocity = √5² + 3² + 8² = √98

acceleration = dV/dt  = 2i + 0j  + (12t - 8)k

at t = 2

acceleration = 2i + 16k

Magnitude of acceleration = √2² + 16² = √260 = 2√65