Question

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration = (4.80 m/s2)î + (7.00 m/s2)ĵ. at time t = 0, the velocity is (4.4 m/s)î. what are magnitude and angle of its velocity when it has been displaced by 12.4 m parallel to the x axis?

Answer 1

Acceleration, a = ( 4.80 i+7.00 j )
`m/s^2`

We have acceleration a = Horizontal component i + Vertical component j

So Horizontal component =4.80
`m/s^2`

Vertical component =7.00
`m/s^2`

We have horizontal displacement ,
`s = ut+(1)/(2) at^2`

Here s = 12.4 m, u = 4.4 m/s and a = 4.80
`m/s^2`

Substituting

`12.4 = 4.4*t+(1)/(2) *4.80*t^2\\ \\ t^2+1.83*t-5.167=0\\ \\ (t-1.53)(t+3.36)=0\\ \\ time = 1.53 seconds`

Negative time is not possible.

Now we need to find velocity after 1.53 seconds

We have v = u+at

Horizontal case

v = 4.4 + 4.80 * 1.53 = 11.744 i

Vertical case

v = 0 + 7.00 * 1.53 = 10.71 j

So Velocity when it has been displaced by 12.4 m parallel to the x axis = (11.744 i+10.71 j) m/s

Magnitude =
`√(11.744^2+10.71^2) =15.89 m/s`

Angle , tan θ =
`(10.71)/(11.744)`

θ = 42.36⁰