1. A runner drops her phone as she is running at a constant speed of 3 miles per hour from point A to point B in a park. Describe the motion of the phone as it is observed by someone sitting on a bench at the park.

Let's define the point A as our zero in the x-axis.

As the phone drops, it keeps the horizontal velocity that it had before, so the horizontal velocity is:

Vx = 3 mi/h.

Now, the only force acting on the phone is the gravitational force that acts in the vertical axis, then we have:

Ay = -g

where g = 9.8 m/s^2

It is dropped, so we do not have a vertical initial velocity, then for the vertical velocity we should integrate over time:

Vy = -g*t

And for the position again, we integrate over time, but now we have an initial position H, that is the height at which the phone is dropped.

Py = -(1/2)*g*t^2 + H

And the horizontal position can be found by integrating over time the horizontal velocity.

Px = (3mi/h)*t

This will be the two equations that describe the motion of the phone, and we can not solve it further because we do not know the initial height of the phone.

But in general, we have a linear equation in the horizontal axis and a quadratic equation with a negative leading coefficient in the vertical axis.

Position(t) = ( (3mi/h)*t,  -(1/2)*g*t^2 + H)

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joke:

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total distance/total time

(85 + 82)/(5+3)

167/8

20 just over