Sandra
0

Given that

Position of the particle r = 3.0t \hat{i} - 2 t^2 \hat{j} + 4.0 \hat{k}

a) Velocity and acceleration

Velocity: change in position with respect to time  

v = \frac{dr}{dt}

v = \frac{d}{dt}(3.0t \hat{i} - 2 t^2 \hat{j} + 4.0 \hat{k})

v = 3.0 \hat{i} - 4t \hat{j} + 0

Hence, velocity of the particle v = 3.0 \hat{i} - 4t \hat{j} m/s

Acceleration: Change in velocity with respect to time

a = \frac{dv}{dt}

a = \frac{d}{dt}(3.0 \hat{i} - 4t \hat{j})

a = - 4 \hat{j} m/s^2

b)

Time t = 2.0 sec

Velocity of the particle v = 3.0 \hat{i} - 4t \hat{j} m/s

v = 3.0 \hat{i} - 4(2) \hat{j} = 3.0 \hat{i} - 8t \hat{j}

Horizontal component v_x = 3.0

Vertical component v_y = - 8   

Magnitude of the velocity |v| = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + (-8)^2}

|v| = \sqrt{9 + 64} = \sqrt{73} = 8.54

Magnitude of the velocity  |v| = 8.54 m/s

Direction of the velocity \tan \theta = \frac{v_y}{v_x}

\tan \theta = \frac{8}{3} =

\theta = \tan^{-1} (2.66)

\theta = - 69.45\degree

Thus, Direction of the velocity \theta = - 69.45\degree