The Problem:
A train is slowing down with constant deceleration. It passes a signal at A, and after successive intervals of 40 seconds it passes points B and C, where AB = 1800 m and BC= 1400 m. (a) How fast is the train moving when it passes A? (b) How far from A does it come to a stop?
Douglas Quadling Mechanics1 Exercise1D Q9
A train is slowing down with constant deceleration. It passes a signal at A, and after successive intervals of 40 seconds it passes points B and C, where AB = 1800 m and BC= 1400 m………
:
Problem: A train is slowing down with constant deceleration. It passes a signal at point A, and after successive intervals of 40 seconds, it passes points B and C, where AB = 1800 meters and BC = 1400 meters.
Questions:
- How fast is the train moving when it passes A?
- How far from A does it come to a stop?
Solution:
Step 1: Understand the Problem
The train is slowing down with constant deceleration. It passes point A, then 40 seconds later, it passes point B, which is 1800 meters from A. Another 40 seconds after passing B, it passes point C, which is 1400 meters from B.
Step 2: Define Variables
- Let u be the speed of the train when it passes A.
- Let a be the constant deceleration (negative value).
Step 3: Use Equations of Motion
We use the equation: s = ut + (1/2)at², where:
- s = distance,
- u = initial speed,
- t = time,
- a = acceleration (deceleration in this case).
From A to B:
- Distance (s) = 1800 meters,
- Time (t) = 40 seconds.
Plugging into the equation:
1800 = u * 40 + (1/2) * a * (40)²
1800 = 40u + 800a
This is Equation 1: 40u + 800a = 1800
From B to C:
- Distance (s) = 1400 meters,
- Time (t) = 40 seconds.
First, find the speed at B (v_B):
v_B = u + a * 40
Now, use the equation for B to C:
1400 = v_B * 40 + (1/2) * a * (40)²
Substitute v_B = u + 40a:
1400 = (u + 40a) * 40 + 800a
1400 = 40u + 1600a + 800a
1400 = 40u + 2400a
This is Equation 2: 40u + 2400a = 1400
Step 4: Solve the Equations
Subtract Equation 1 from Equation 2:
(40u + 2400a) – (40u + 800a) = 1400 – 1800
1600a = -400
a = -400 / 1600
a = -0.25 m/s²
Now, substitute a = -0.25 into Equation 1:
40u + 800 * (-0.25) = 1800
40u – 200 = 1800
40u = 2000
u = 2000 / 40
u = 50 m/s
Answer to (a): The speed of the train when it passes A is 50 m/s.
Step 5: Find Distance from A to Stop
When the train stops, its final speed v = 0. Use the equation:
v² = u² + 2as
Plugging in the values:
0 = (50)² + 2 * (-0.25) * s
0 = 2500 – 0.5s
0.5s = 2500
s = 2500 / 0.5
s = 5000 meters
Answer to (b): The train comes to a stop 5000 meters from A.
Final Answers:
- The speed of the train when it passes A is 50 m/s.
- The train comes to a stop 5000 meters from A.
