A train has a mass of 5.22 x10^6 kg and is moving at 96.0 km/h.?

Question

The engineer applies the brakes, resulting in a net backward force of 1.87x10^6 N on the train. The brakes are held on for 33.0 s.

a)What is the final speed of the train? ___m/s

b)How far does the train travel during this period?_____m

need help on this one. thanks

Answer 1

a)F=ma
(1.87x10^6N)=(-5.22x10^6)a
a=-.358m/s^2
vf=vi+at
have to convert km/hr to m/s by multiplying by 10/36
vf=26.67+(-.358)(33)
vf=14.84m/s

b)vf^2=vi^2+2as
(14.84)^2=(26.67)^2+2(-.358)s
s=685.4m

Answer 2

Easy enough. Use the first law of motion, F = ma. The force on the train is -1.87x10^6 N; the mass is 5.22x10^6 kg. So, the acceleration is therefore -0.36 m/s^2. The train is moving at 96 km/h, or 26.67 m/s. If it is decelerating for 33 s, its final speed will be 26.67 - 0.36 * 33 m/s, or 14.85 m/s. How far did it travel? There's a standard formula to compute this for a system under acceleration, which you hopefully know. It's worth memorizing if you don't know how to derive it from first principles: x = v_0 * t + 1/2 * a * t^2, where v_0 is the initial velocity and x is the distance traveled. In our case this comes out to 1246 m.

Answer 3

a) a = F/m = -1.87E6/5.22E6 = -.358 m/s² ∆V = a*t = -11.821 m/s
......Vf = Vo + ∆V = 96 - (11.821*3600/1000) = 53.44 km/hr

b) D = Vo*t + .5a*t² = 26.667*33 - .5*.358*33² = 685.1 m