Question

Freight trains can produce only relatively small accelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500m/s2 for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550m/s2, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?

Answer 1

Answer:

a) 28 m/s   b) 50.9s   c)i)7680m   c)ii) 712m

Explanation:

The acceleration is constant in each case, so lets use the ecuations for this kind of motion.

a) For the velocity we have:

$v= v_{0} +at= 4 + 0.05000\times480=28\frac{m}{s}$

where 480 is the amount of seconds in 8 min

$480 = 60\times8$

b) In this case we got a 0 velocity, starting with 28 m/s and accelerating at a rate of -0550m/s2 , that means , slowing down.

$0=v= v_{0} +at= 28 - 0.550t$

Solving for t:

$t=\frac{28}{0.55} = 50.9s$

c) Now we can use the equation for the position of a particle with constant acceleration.

$x=x_{0} + v_{0}t+\frac{1}{2} at^{2}$

In the case of a) we have:

$x=0 + 4\times480+\frac{1}{2}\times 0.0500\times (480)^{2} =7680m$

In the case of b) we have

$x=0 + 28\times50.9+\frac{1}{2}\times -0.550\times (50.9)^{2} =712m$

Hope my answer helps you.

Have a nice day!