161 km/h = 44.722 m/s
29 km/hr = 8.056 m/s
Now just solve the usual equations.
The distance traveled by the train must equal the distance traveled by the locomotive, plus the 676m separating them at first:
44.722 + at = 0
44.722t + 1/2 at^2 = 676 + 8.056t
a = -0.946 m/s^2
When a high-speed passenger train traveling at 161 km/h rounds a bend,the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding
and is a distance(D)= 676m ahead.The locomotive is moving at 29.0 km/h.The engineer of the high-speed train immediately applies the brakes.What must be the magnitude of the
resulting constant deceleration if a collision is to be just avoided
Solve
1 answer
1 answer