Question

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration ModifyingAbove a With right-arrow equals left-parenthesis 5.0ModifyingAbove i With caret plus 8.0ModifyingAbove j With caret right-parenthesis m/s squared. At time t = 0, the velocity is left-parenthesis 5.0ModifyingAbove i With caret right-parenthesis m/s. What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

Answer 1

The magnitude of the pebble's velocity is 20.0 m/s and the angle of the velocity is 57.99 degrees.

Step-by-step explanation:

To find the magnitude and angle of the pebble's velocity after it has been displaced by 11.0 m parallel to the x-axis, we can use the equations of motion. First, we need to find the time it takes for the pebble to be displaced by 11.0 m. We can use the equation for displacement: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Rearranging the equation gives us t = (2s - u) / a. Plugging in the values, we get:

t = (2 * 11.0 - 5.0) / 5.0 = 3.2 s

Now, we can find the final velocity using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get:

v = 5.0 + (5.0 * 3.2) = 20.0 m/s

The magnitude of the velocity is 20.0 m/s. To find the angle of the velocity, we can use the tangent of the angle, which is equal to the y-component of the velocity divided by the x-component. The y-component is 8.0 m/s and the x-component is 5.0 m/s, so the angle is atan(8.0/5.0) = 57.99 degrees.

Answer 2

To answer this problem we will make use of two of the equations of motion. Then we will come across out each velocity component. The difficult part comes in solving for the j component of velocity for which we will have to know t_f, the time it takes to be displaced (11m):

Our equations
1. v_f^2 = v_0^2 + 2ax

2. v_f = v_0 + at


Solving for the time it takes to travel 11m

We do so by finding the final velocity using (1), and then plugging that back into (2).

v_fi^2 = 4^2 + 2*5*11

v_fi^2 = 16 + 110 = 126

v_fi = 11.22 m/s ((()))

plugging this into (2)

11.22 = 4 + 5*t_f

5*t _f= 7.22

t_f = 7.22 / 5

t_f = 1.444 seconds

Solving for the vertical velocity

v_fj = v_0j + at

v_fj = 0 + 7*1.444

v_fj = 10.108 m/s ((()))

Finding magnitude and angle

V = sqrt(v_fi^2 + v_fj^2)

125.884 + 10.108

V = 15.10151 m/s

Angle:

Theta = arctan(v_fj/v_fi)

Theta = 0.0801 radians

converting radians to degrees (180/pi)

Theta = .0.0801(180/pi) deg= 4.591 deg

Answer:

V = 15.10151 m/s

Theta = 4.591 deg