Exam Questions - Velocity time graphs - ExamSolutions - Free Printable

Problem Analysis

We are given a velocity-time graph for a train \( P \) traveling between two stations 1.5 km apart. The train accelerates uniformly to a speed of 30 m/s over the first 300 meters, maintains this speed for \( T \) seconds, and then decelerates uniformly at \( 1.25 \, \text{m/s}^2 \) until it comes to rest. We need to solve for various parameters related to this journey and compare it with another train \( Q \).

#### Given:
- Distance between stations: \( 1.5 \, \text{km} = 1500 \, \text{m} \)
- Initial speed: \( u = 0 \, \text{m/s} \)
- Final speed after acceleration: \( v = 30 \, \text{m/s} \)
- Distance covered during acceleration: \( s_1 = 300 \, \text{m} \)
- Deceleration rate: \( a_{\text{decel}} = -1.25 \, \text{m/s}^2 \)

(a) Find the acceleration of \( P \) during the first 300 m of its journey.

Step 1: Use the kinematic equation for uniform acceleration.
The equation relating distance, initial velocity, final velocity, and acceleration is:
\[
v^2 = u^2 + 2as
\]
where:
- \( v = 30 \, \text{m/s} \)
- \( u = 0 \, \text{m/s} \)
- \( s = 300 \, \text{m} \)
- \( a \) is the acceleration we need to find.

Substitute the known values:
\[
30^2 = 0^2 + 2a(300)
\]
\[
900 = 600a
\]
\[
a = \frac{900}{600} = 1.5 \, \text{m/s}^2
\]

Answer:
\[
\boxed{1.5}
\]

(b) Find the value of \( T \).

Step 1: Calculate the time taken to accelerate to 30 m/s.
Using the equation:
\[
v = u + at
\]
where:
- \( v = 30 \, \text{m/s} \)
- \( u = 0 \, \text{m/s} \)
- \( a = 1.5 \, \text{m/s}^2 \)

Substitute the known values:
\[
30 = 0 + 1.5t
\]
\[
t = \frac{30}{1.5} = 20 \, \text{s}
\]

So, the time to accelerate is \( 20 \, \text{s} \).

Step 2: Calculate the distance covered during deceleration.
Using the equation:
\[
v^2 = u^2 + 2as
\]
where:
- \( v = 0 \, \text{m/s} \) (final speed at rest)
- \( u = 30 \, \text{m/s} \)
- \( a = -1.25 \, \text{m/s}^2 \)

Substitute the known values:
\[
0^2 = 30^2 + 2(-1.25)s
\]
\[
0 = 900 - 2.5s
\]
\[
2.5s = 900
\]
\[
s = \frac{900}{2.5} = 360 \, \text{m}
\]

So, the distance covered during deceleration is \( 360 \, \text{m} \).

Step 3: Calculate the distance covered at constant speed.
The total distance between the stations is \( 1500 \, \text{m} \). The distance covered during acceleration and deceleration is:
\[
300 \, \text{m} + 360 \, \text{m} = 660 \, \text{m}
\]
Thus, the distance covered at constant speed is:
\[
1500 \, \text{m} - 660 \, \text{m} = 840 \, \text{m}
\]

Step 4: Calculate the time \( T \) spent at constant speed.
Using the equation:
\[
s = vt
\]
where:
- \( s = 840 \, \text{m} \)
- \( v = 30 \, \text{m/s} \)

Substitute the known values:
\[
840 = 30T
\]
\[
T = \frac{840}{30} = 28 \, \text{s}
\]

Answer:
\[
\boxed{28}
\]

(c) Sketch on the diagram above a velocity-time graph which represents the journey of train \( Q \).

Train \( Q \) accelerates uniformly from rest to a speed \( V \, \text{m/s} \) and then decelerates uniformly to rest. The total distance covered is still 1500 meters, and the total time is the same as for train \( P \).

1. Acceleration phase: The train accelerates from \( 0 \, \text{m/s} \) to \( V \, \text{m/s} \).
2. Deceleration phase: The train decelerates from \( V \, \text{m/s} \) to \( 0 \, \text{m/s} \).

The velocity-time graph will be a straight line rising from 0 to \( V \), followed by a straight line falling back to 0. The area under this graph must equal 1500 meters, and the total time must match the total time for train \( P \).

(d) Find the value of \( V \).

Step 1: Calculate the total time for train \( P \).
The total time for train \( P \) is the sum of the time spent accelerating, the time spent at constant speed, and the time spent decelerating:
\[
t_{\text{total}} = t_{\text{accel}} + T + t_{\text{decel}}
\]
We already know:
\[
t_{\text{accel}} = 20 \, \text{s}, \quad T = 28 \, \text{s}
\]

Calculate the time spent decelerating:
Using the equation:
\[
v = u + at
\]
where:
- \( v = 0 \, \text{m/s} \)
- \( u = 30 \, \text{m/s} \)
- \( a = -1.25 \, \text{m/s}^2 \)

Substitute the known values:
\[
0 = 30 + (-1.25)t
\]
\[
0 = 30 - 1.25t
\]
\[
1.25t = 30
\]
\[
t = \frac{30}{1.25} = 24 \, \text{s}
\]

So, the total time for train \( P \) is:
\[
t_{\text{total}} = 20 + 28 + 24 = 72 \, \text{s}
\]

Step 2: Use the total time and distance for train \( Q \).
For train \( Q \), the total time is also 72 seconds, and the total distance is 1500 meters. The velocity-time graph is a triangle, so the area under the graph is:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 72 \times V = 1500
\]

Solve for \( V \):
\[
\frac{1}{2} \times 72 \times V = 1500
\]
\[
36V = 1500
\]
\[
V = \frac{1500}{36} = \frac{125}{3} \approx 41.67 \, \text{m/s}
\]

Answer:
\[
\boxed{\frac{125}{3}}
\]