Time needed to accelerate a freight train?
First calculate the acceleration: Mass=2.230,000 kg force =43,400 N F=MA=>A=F/M=43,400 N/2,230,000 kg=.0194 m/s^2 Then convert velocity from km/hr to m/s: 48.2 km/hr==48,200 m/3600 s=13.39 m/s Now calculate the time needed to accelerate to given velocity: Acceleration x Time=Velocity ==>Time =Velocity/acceleration (13.39 m/s)/(.0194 m/s^2)=690.14 S
If a car accelerates at 4.0 m/s2, how long will it take to reach a speed of 80 km/hr, starting from rest?
You need to look at the units of your values and do the right cancellation to end up with what you need. So lets see if we can do that here: I would start by converting the 80km/hr into units of m/s so we can make the proper comparison to the acceleration units. What is 80 km in m? 80 km x 1000m/1km = 80,000 m What is 1 hr in s? 1 hr x 60 min/1 hr x 60 s/1 min = 3600 s So the speed we need to reach is 80,000 m / 3600 s = 22.22 m/s Now, what does it mean to accelerate? It means that we increase our speed by a certain number per unit time. We say it is the rate of change of velocity over time. We are talking about UNIFORM or constant acceleration which is a type of motion in which the velocity of an object changes by an equal amount in every equal time period. That's why our units are meters per second per second. So in your problem, we know we are increasing our speed by 4.0 m/s every second. So the question we ask ourselves is how many 4.0 values can go into the final speed of 22.22? This makes sense because this is the number were are trying to REACH. Using division we get, 22.22 m s^-1/4.0 m s^-2 = 5.6 seconds So if we start from stop.....and accelerate 4.0 m per second per second, it would take 5.6 seconds to reach a speed of 22.22 meters per second (80 km/hr).
Science problems with calculating speed and acceleration?
speed is the cost at which place is changing, so speed could be chanced on by using dividing the exchange in distance by using the exchange in time, or in calculus words, dx/dt. Acceleration is that value at which the speed is changing, so acceleration could be chanced on by using dividing the exchange in speed by using the exchange in time, or in calculus words, dv/dt. in case you're attentive to calculus, in case you're given a place function, the 1st by-product provide you a function instant speed and the 2nd by-product provide you a function of instanenous acceleration. in case you're given a speed function, the acceleration function may be the by-made of it. in case you're given acceleration, speed may be the anti-by-made of that. desire I helped.
How long does it take to stop a train?
It's really hard to answer that as too many things can affect how fast you can stop.Weight and grade plays the biggest factor,but different types of cars stop better than others too.Passenger trains are very light so they stop pretty quick.I run freight so as a guess i'd say a passenger train would stop in 3/4's to 1 mile.A heavy freight like a coal train(15,000 tons or more) can take upwards of 2 miles.The average freight train,about 1 1/2 miles.That is for stops in what the railroad calls consistent with good train handling(meaning not having to do a panic stop).As for getting back to speed again it all depends on how heavy you are and how much power you have.A passenger train would take about 2-3 minutes to get back to speed. Edit With all your variables i'll stick with my 3/4's of a mile.Passengers get real cranky when you spill their coffee.Rango might be closer though.I know they stop pretty quick.Passenger trains also have the ability of easing off the brakes if they need too.A freight train uses a different system of braking.Once you set the brakes on a freight your stuck with how much air you set unless you release it and start over.
Acceleration question- answer given but need explanation?
The average of any two values x & y (assuming a linear relationship, which you have here since it's CONSTANT acceleration) is Average = (x + y)/2 In your case it's Average speed = (initial speed + final speed)/2 So just plug in what you know 21 mph = (6 mph + final speed)/2 2 * 21 mph = 6 mph + final speed 2 * 21 mph - 6 mph = final speed 42 mph - 6 mph = final speed final speed = 36 mph I'm using speed not velocity since there is no direction component
How long does the average freight train take to pass?
It depends on length and speed. A train one mile long would take: 45 seconds at 80 MPH 48 seconds at 75 MPH 56 seconds at 65 MPH 1 minute at 60 MPH. 1 minute and 6 seconds for 55 MPH 1 minute and 12 seconds at 50 MPH 1 minute and 20 seconds at 45 MPH 1 minute and 30 seconds at 40 MPH 1 minute and 43 seconds at 35 MPH 2 minutes at 30 MPH 2 minutes and 26 seconds at 25 MPH 4 minutes at 15 MPH 6 Minutes at 10 MPH If a train were 1 and 1/2 miles long, then you would add half again as much, etc. It is definitely worth waiting for these trains to pass if your intention is to reach a ripe old age...................