Question

The velocity of a 3.00 kg parti- cle is given by :v = (8.00tiˆ + 3.00t2jˆ) m/s, with time t in seconds. At the instant the net force on the parti- cle has a magnitude of 35.0 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer 1

Part a)

Direction of net force is

`\theta = 46.7 degree`

Part b)

Direction of the velocity is given as

`\phi = 27.9 degree`

Step-by-step explanation:

As we know that the velocity of the particle is given as

`v = 8.00 t \hat i + 3.00 t^2\hat j`

now the acceleration is given as

`a = (dv)/(dt)`

`a = 8.00 \hat i + 6.00 t\hat j`

now magnitude of net acceleration is given as

`a = √(64 + 36t^2)`

`F = 3a`

`35 = 3√(64 + 36t^2)`

`t = 1.41 s`

Part a)

Now direction of net force is given as

`tan\theta = (F_y)/(F_x)`

`tan\theta = (6t)/(8)`

`tan\theta = (6(1.41))/(8)`

`\theta = 46.7 degree`

Part b)

Direction of the velocity is given as

`tan\phi = (v_y)/(v_x)`

`tan\phi = (3.00 t^2)/(8.00 t)`

`tan\phi = (3.00(1.41))/(8.00)`

`\phi = 27.9 degree`