Time for freight train to accelerate from rest to 85 km/h

[mike98020205]

Help, I just can't seem to get this.

A freight train has a mass of 15000000 kg. If the locomotive can exert a constant pull (what is a constant pull?) of 750000 N, how long would it take to increase the speed of the train from rest to 85 km/h? (Disregard friction.)

 

[Atomos]

Newtons second law: f=ma, a=f/m. You can use this to derive the acceration and use some kinematics formulas to find out how long it will take to reach a certain velocity with a known acceleration and a known initial velocity.

 

[Poncho]

F = ma

750000 N = (15000000 kg) * a

 

[Giuseppe]

Well, you have a fairly basic question that comes from Netwon's second law of motion. By constant pull, the system has a constant force that is pulling the train along. This means that the system also has constant acceleration, which is given by Netwon's Laws.

F = M A

You can easily find this acceleration by using the values already given.
You now can use this in a kinematics equation. You have acceleration,initial velocity,final velocity, and you need to find time.

I suggest using this equation:

V(f)=V(i)+at

By the way, make sure you watch your units. The final velocity that the problem gives is in kilometers per hour. Convert this speed to standard units.

Hope this helps

 

[mike98020205]

Ok... What is the answer? The book says 470 seconds. I did 750000N/15000000 and got .05m/s squared for the acceleration. Then, If I divide the final speed, 85 km/h by this acceleration, I get 1700 seconds which is wrong. How do they get 4.7 times 10 to the second (470 seconds)? Help please. I appreciate it.

 

[Atomos]

conversion from m/s to km/h

 

[mike98020205]

They are looking for time in this problem.

 

[Atomos]

yeah, but the conversion from m/s to km/h is essential in arriving at the proper time. You are dividing 85 km/h by m/s. The units do not cancel out.