Question

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 mm????????2. 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?

Answer 1

You have to run 73.8 m at a speed of 11.9 m/s

Step-by-step explanation:

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

If the object has no acceleration, then, a = 0 and x = x0 + v · t, where v is the constant velocity.

When you catch the rear of the bus, its position and yours will be the same:

your position = position of the bus

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

since you start from rest and the origin of the reference system is located at the point where you start running, x0 and v0 = 0.

The initial position of the bus will be 12.0 m because this was its position relative to you when you started running. Then:

1/2 · 0.960 m/s² · t² = 12.0 m + 5.00 m/s · t

0.480 m/s² · t² - 5.00 m/s · t - 12.0 m = 0

solving this quadratic equation:

t = 12.4 s (The other solution is negative and therefore discarded)

Now, with this time, we can calculate your position:

x = 1/2 · a · t²

x = 1/2 · 0.960 m/s² · (12.4 s)² = 73.8 m

Your speed can be calculated with the equation for speed:

v = v0 + a · t

Since v0 = 0

v = a · t

v = 0.960 m/s² · 12.4 s = 11.9 m/s (really fast!)

Answer 2

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

Step-by-step 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:

`X_y=(1)/(2) (0.960(mm)/(s^(2)))*t^(2)`

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

`X_b=X_y\\12+5t=(1)/(2) 0.960t^(2) \\0.48t^(2)-5t-12=0`

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:

`X_y=(1)/(2) (0.960(mm)/(s^(2)))*12.4282^(2)`

Xy = 74.1409 mm

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

`Vy=0.960(mm)/(s^(2) )*t`

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

`Vy=0.960(mm)/(s^(2) )*12.4282s`

Vy = 11.9311 mm/s