Acceleration of a train
The train passes 700 m, braking with an acceleration of -0.15 m/s2. How long does it break, and what is the final speed of the train if the initial was 55 km/h?
Correct answer:
Showing 1 comment:
Dr. Math
To solve this problem, we will use the kinematic equations for uniformly accelerated motion. The train is braking, so the acceleration is negative ( a = -0.15 m/s2 ). We are tasked with finding:
1. The time it takes for the train to brake ( t ).
2. The final speed of the train ( v ).
Given Data:
1. Initial speed of the train, u = 55 km/h .
2. Braking acceleration, a = -0.15 m/s2 .
3. Distance traveled while braking, s = 700 m .
Step 1: Convert Initial Speed to m/s
The initial speed is given in km/h, but we need it in m/s for consistency with the acceleration units. Convert 55 km/h to m/s:
Step 2: Use the Kinematic Equation to Find Final Speed
The kinematic equation that relates initial velocity ( u ), final velocity ( v ), acceleration ( a ), and distance ( s ) is:
Substitute the known values ( u = 15.28 m/s , a = -0.15 m/s2 , s = 700 m ):
Calculate u2 :
Calculate 2as :
Now, substitute these values into the equation:
Take the square root to find v :
Step 3: Convert Final Speed to km/h
The final speed is in m/s, but it is often more intuitive to express it in km/h. Convert 4.85 m/s to km/h:
Step 4: Use the Kinematic Equation to Find Time
The kinematic equation that relates initial velocity ( u ), final velocity ( v ), acceleration ( a ), and time ( t ) is:
Solve for t :
Substitute the known values ( v = 4.85 m/s , u = 15.28 m/s , a = -0.15 m/s2 ):
Calculate the numerator:
Now, divide by a :
Final Answers:
1. The train brakes for:
2. The final speed of the train is:
1. The time it takes for the train to brake ( t ).
2. The final speed of the train ( v ).
Given Data:
1. Initial speed of the train, u = 55 km/h .
2. Braking acceleration, a = -0.15 m/s2 .
3. Distance traveled while braking, s = 700 m .
Step 1: Convert Initial Speed to m/s
The initial speed is given in km/h, but we need it in m/s for consistency with the acceleration units. Convert 55 km/h to m/s:
u = 55 × 1000/3600 = 15.28 m/s.
Step 2: Use the Kinematic Equation to Find Final Speed
The kinematic equation that relates initial velocity ( u ), final velocity ( v ), acceleration ( a ), and distance ( s ) is:
v2 = u2 + 2as.
Substitute the known values ( u = 15.28 m/s , a = -0.15 m/s2 , s = 700 m ):
v2 = (15.28)2 + 2(-0.15)(700).
Calculate u2 :
(15.28)2 = 233.48.
Calculate 2as :
2(-0.15)(700) = -210.
Now, substitute these values into the equation:
v2 = 233.48 - 210 = 23.48.
Take the square root to find v :
v = √23.48 = 4.85 m/s.
Step 3: Convert Final Speed to km/h
The final speed is in m/s, but it is often more intuitive to express it in km/h. Convert 4.85 m/s to km/h:
v = 4.85 × 3600/1000 = 17.46 km/h.
Step 4: Use the Kinematic Equation to Find Time
The kinematic equation that relates initial velocity ( u ), final velocity ( v ), acceleration ( a ), and time ( t ) is:
v = u + at.
Solve for t :
t = v - u/a.
Substitute the known values ( v = 4.85 m/s , u = 15.28 m/s , a = -0.15 m/s2 ):
t = 4.85 - 15.28/-0.15.
Calculate the numerator:
4.85 - 15.28 = -10.43.
Now, divide by a :
t = -10.43/-0.15 = 69.53 seconds.
Final Answers:
1. The train brakes for:
69.53 seconds
2. The final speed of the train is:
17.46 km/h
Tips for related online calculators
Are you looking for help with calculating roots of a quadratic equation?
Need help calculating sum, simplifying, or multiplying fractions? Try our fraction calculator.
Do you have a linear equation or system of equations and are looking for its solution? Or do you have a quadratic equation?
Do you want to convert velocity (speed) units?
Do you want to convert time units like minutes to seconds?
Need help calculating sum, simplifying, or multiplying fractions? Try our fraction calculator.
Do you have a linear equation or system of equations and are looking for its solution? Or do you have a quadratic equation?
Do you want to convert velocity (speed) units?
Do you want to convert time units like minutes to seconds?
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Themes, topics:
Grade of the word problem:
We encourage you to watch this tutorial video on this math problem: video1
Related math problems and questions:
- Acceleration 7054
The car brakes with an acceleration of 5 m/s-2. Determine the stopping distance of the car if its initial speed is a) 54 km/h, b) 108 km/h. Compare the calculated braking distances concerning the given speeds. - Acceleration 4211
At the start of braking, the car had a speed of 72 km h at -1. It stopped on a track of 50 m. What was the acceleration, and how long did the braking last? - Brakes
The braking efficiency of a passenger car is required to stop at 12.5 m at an initial speed of 40 km/h. What is the acceleration braking by brakes? - Acceleration 80626
The car stopped at an initial speed of 90 km/h on a track of 62.5 m. What was its acceleration, and how long did it take to brake the car?
- Acceleration 7123 The car starts to brake with an acceleration of -6.5 m/s² and stops on the track at 45m. What were the initial speed and braking time?
- Subway
The Subway train went between two stations that gradually accelerated for 26 seconds and reached a speed of 72 km/h. At this rate, it went 56 seconds. Then 16 seconds slowed to a stop. What was the distance between the stations? - Train
Find the average speed of a train (in km/h) that passes a telegraph pole in 7 seconds and passes a 378 m platform in 25 seconds. How long is the train?
