# A train has a mass of 6.32x104 kg and is moving with a speed of 94.3 km/hr. The engineer applies the brakes which results in a net backward force of 2.43x105 Newtons on the train. The brakes are held for 3.40 seconds. How far (in meters) does the train travel during this time?

Mukesh Sharma
10 years ago
There are three (perhaps four) forces acting upon this train. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding train. In the free-body diagram, these two forces are represented by the Ffrict arrow. The value of this resistive force is given as 2.43x105 N. This is the net force since there are no other horizontal forces; it is the force which causes the acceleration of the train. The acceleration value can be determined using Newton's second law of motion.
a = Fnet / m = (2.43x105 N) / (6.32x104 kg) = 3.84 m/s/s, left

This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds.

d = vi • t + 0.5 •a • t2

d = (26.2 m/s) • (3.40 s) + 0.5 •(-3.84 m/s/s) • (3.40 s)2

d = 89.1 m - 22.2 m

d = 66.8 m

Mukesh Sharma
10 years ago
There are three (perhaps four) forces acting upon this train. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding train. In the free-body diagram, these two forces are represented by the Ffrict arrow. The value of this resistive force is given as 2.43x105 N. This is the net force since there are no other horizontal forces; it is the force which causes the acceleration of the train. The acceleration value can be determined using Newton's second law of motion.
a = Fnet / m = (2.43x105 N) / (6.32x104 kg) = 3.84 m/s/s, left
This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds.
d = vi • t + 0.5 •a • t2
d = (26.2 m/s) • (3.40 s) + 0.5 • (-3.84 m/s/s) • (3.40 s)2
d = 89.1 m - 22.2 m
d = 66.8 m
Thanks & Regards
Mukesh Sharma