Question

When a high-speed passenger train traveling at vP = 153 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 = 847 m ahead (see the figure). The locomotive is moving at vL = 28 km/h. The engineer of the passenger train immediately applies the brakes. Assume that an x axis extends in the direction of motion. What must be the constant acceleration along that axis if a collision is to be just avoided?

Answer 1

The forward distance travelled during this time by the locomotive is

s= v_l t

trains velocity vt

locomotive velocity v_l

del x is gap between train and locomotive

vt+ v_l/2 = del x/ t= D+ v_l t/ t = D/t + v_l

vt+ v_l/2 =D/( v_l - vt)/a + v_l

a= vt+ v_l/2 - v_l) ( v_l - vt/D)

= - 1/2D * ( v_l- vt)^2

= - 1/ 2(0.847 km) * ( 28 km/h - 153 km/h)^2

=9223.73 km/h^2

a= - 9223.73 km/h^2 ( 1000 m/km)( 1h/ 3600s)^2

=-0.711 m/s^2

The magnitude of accleration is

a= 0.711 m/s^2