Answer:
26m/s
Explanation:
Assuming that the acceleration is constant, we can start by calculating the train speed when it's free of the congested area:
![a = \frac{\deltav}{\deltat} = \frac{12 - 5}{8} = \frac{7}{8} = 0.875 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5Cdeltav%7D%7B%5Cdeltat%7D%20%3D%20%5Cfrac%7B12%20-%205%7D%7B8%7D%20%3D%20%5Cfrac%7B7%7D%7B8%7D%20%3D%200.875%20m%2Fs%5E2)
Then with the same acceleration we can find out the final speed:
![v = v_0 + at = 12 + 0.875*16 = 26m/s](https://tex.z-dn.net/?f=v%20%3D%20v_0%20%2B%20at%20%3D%2012%20%2B%200.875%2A16%20%3D%2026m%2Fs)
Answer:
if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.
Y (position)
I
sensor-------------------------------------------------------------------
I o
I o
I o
I o
I o
I o
I o
I------------------------------------------------------------------------------------- X (time)
in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.
Y (position)
I
sensor ----------------------
I o
I o
I o
I o
I o
I o
I o
I------------------------------------------------------------------------------------- X (time)
i hope you can understand it, kinda hard to do graphs here.
100% True, as gravity has a force on everything in the universe