The correct answer to your question here is D
Answer:
The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24
.
Explanation:
Velocity is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it. Then, the average velocity relates the change in position to the time taken to effect that change.

Velocity considers the direction in which an object moves, so it is considered a vector magnitude.
In this case, the displacement is 192 m and the time period is 8 s. Replacing:

Solving:
velocity= 24 
<em><u>The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 </u></em>
<em><u>.</u></em>
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m