Question

Amtrak's 20th-Century Limited is en route from Chicago to New York at 100 km/h when the engineer spots a cow on the track. The train brakes to a halt in 1.6 min , stopping just in front of the cow.

a) What is the magnitude of the train's acceleration?

b)What's the direction of the acceleration? eastward or westward?

c)How far was the train from the cow when the engineer first applied the brakes?

Answer 1

initial velocity u =100 Km/hr

u = 100 X 1000/3600

u = 27.78 m/s

t = 1.6 min = 1.6 x 60

t = 96 s

final velocity v = 0

from first equation of motion

v = u + at

0 = 27.78 + a(96)

96a = -27.78

a= -27.78/96

a = -0.29 m/s²

a) The magnitude of acceleration is 0.29 m/s²

b) The acceleration is negative which means it is in the direction opposite to the direction of motion of train i.e acceleration is directed from New york to Chicago

c) To calculate the distance travelled by train just before stopping we have

initial velocity u = 27.78 m/s

acceleration a = -0.29

time t = 96s

Distance S can be calculated using second equation of motion

S = ut + (1/2)at²

S = (27.78)(96) + (0.5)(-0.29)(96)²

S = 2666.8 - 1336.3

S = 1330.5 m

S = 1.33 Km

The train was 1.33 Km away from crow when Engineer applied the brakes