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The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 m/s^2 in a direction opposite to the train's velocity, while the freight train continues with constant speed.Take x=0 at the location of the front of the passenger train when the engineer applies the brakes.

(a) Will the cows nearby witness a collision?

(b) If so, where will it take place?

(c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

(d) If the speed of the two trains and the deceleration of the passenger train remain as originally stated, how much of a head start would the freight train need in order to avoid a collision?

(e) If instead, the speed of the two trains and the head start of the freight train remain as originally stated, find the deceleration of the passenger train necessary to avoid a collision.

(f) Suppose that someone on board the freight train got the message that there was a problem and begin to accelerate to escape the collision. How much acceleration would be necessary?  Edit Comment  Solutions
-1

a)

when the brake applied, t=0, x = 0, vi = 25.0m/s, a =-0.100 m/s2 for passenger train.

let t be time take to slow the train to 15 m/s, as long as thetrain slow to this speed, it will not collide to the freight

train in the front.

vf =15 m/s = vi + at = 25.0m/s -0.100m/s2*t

=>t = (15m/s- 25m/s) / (-0.100m/s2) = 100.0s

distance travelled during this time is:

d1 = vi t+ (1/2)at2 = 25.0m/s *100s +(1/2)*(-0.100m/s2)*(100s)2 = 2000.0,the train at x = 2000m

Â Â Â Â Â Â Â

freight train at x = 200m, v = 15 m/s, , at t=100s, thedistance travelled is:

d2 = 15m/s*100s = 1500m, the train at x = 1700m

so the collision happened since the passenger train will hit thefreight train before slow down to 15m/s

b)

the collision will happen when d1 = d2 + 200m

d1 = vi t + (1/2)at2

d2 = 15m/s * t

let d1 = d2 + 200m

=>25.0m/s * + 0.5*(-0.100m/s2) t2 =15m/s * t + 200m

=> -0.05 t2 + 10 t - 200m = 0

it will happen at t = 22.54s

d2 = 15m/s*22.54s = 338.1 m

location at x = d2 + 200m = 338.1 + 200m = 538.1m  Edit Comment  Close Choose An Image
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