Answer:
electric motors is the answer
Average acceleration over a time interval lasting
is
![a_{\rm ave}=\dfrac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a_%7B%5Crm%20ave%7D%3D%5Cdfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
where
is the difference in the jet's final and initial velocities. It's coming to a rest, so
![a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}](https://tex.z-dn.net/?f=a_%7B%5Crm%20ave%7D%3D%5Cdfrac%7B0-120%5Cfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D%7B13.5%5C%2C%5Crm%20s%7D%3D-8.9%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D)
so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).
Answer:
![t = 37.6 s](https://tex.z-dn.net/?f=t%20%3D%2037.6%20s)
Explanation:
As we know that train is initially moving with the speed
![v_i = 75.1 km/h](https://tex.z-dn.net/?f=v_i%20%3D%2075.1%20km%2Fh)
now we know that
![v_i = 20.86 m/s](https://tex.z-dn.net/?f=v_i%20%3D%2020.86%20m%2Fs)
now the final speed of the train when it crossed the crossing
![v_f = 15 km/h](https://tex.z-dn.net/?f=v_f%20%3D%2015%20km%2Fh)
![v_f = 4.17 m/s](https://tex.z-dn.net/?f=v_f%20%3D%204.17%20m%2Fs)
now we can use kinematics here
![v_f^2 - v_i^2 = 2 a d](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202%20a%20d)
![4.17^2 - 20.86^2 = 2 a(471)](https://tex.z-dn.net/?f=4.17%5E2%20-%2020.86%5E2%20%3D%202%20a%28471%29)
![a = -0.44 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20-0.44%20m%2Fs%5E2)
Now the time to cross that junction is given as
![v_f - v_i = at](https://tex.z-dn.net/?f=v_f%20-%20v_i%20%3D%20at)
![4.17 - 20.86 = a(-0.44)](https://tex.z-dn.net/?f=4.17%20-%2020.86%20%3D%20a%28-0.44%29)
![t = 37.6 s](https://tex.z-dn.net/?f=t%20%3D%2037.6%20s)