You are standing at rest at a bus stop. a bus moving at a constant speed of 5.00 mm/ passes you. when the rear of the bus is12.0 mm past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 . how far would you have to run before you catch up with the rear of the bus, and how fast must you be running then?

You run 74.1409 mm and you are running at 11.9311 mm/s

Explanation:

If the bus is moving at a constant speed of 5.00mm/s and you start to run when the bus pass you by 12 mm, the equation that describe the position of the bus is:

Xb = 12.0 mm + (5.00 mm/s)*t

Where t is the time in seconds.

If you start to run toward it with a constant acceleration of 0.960 mm/s2, the equation that describe your position is:

So, the time t when you catch up the rear of the bus is the time when Xb is equal to Xy. This is:

Then, solving the quadratic equation, we obtain that t is equal to 12.4282 s

So, if we replace this value of t in the equation of Xy, we obtain how far you have run before you catch up with the rear of the bus. This is:

Xy = 74.1409 mm

Then, the equation of your velocity a time t can be write as:

So, the velocity when you catch up the rear of the bus is:

Vy = 11.9311 mm/s

show the diagram then i will answer

measuring things 1 litre of water