The Problem:
A motor-scooter moves from rest with acceleration 0.1 m s2. Find an expression for its speed, v m s1, after it has gone s metres. Illustrate your answer by sketching an (s, v) graph.
Douglas Quadling Mechanics1 Exercise1C Q
Problem Statement:
A train goes into a tunnel at 20 m/s and emerges from it at 55 m/s. The tunnel is 1500 m long. Assuming constant acceleration, find:
- How long the train is in the tunnel for.
- The acceleration of the train.
Solution:
Step 1: Use the First Equation of Motion
The first equation of motion is:
Final Velocity = Initial Velocity + Acceleration * Time
Substitute the known values:
55 = 20 + a * t
a * t = 55 – 20 = 35 (Equation 1)
Step 2: Use the Second Equation of Motion
The second equation of motion is:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time²
Substitute the known values:
1500 = 20 * t + 0.5 * a * t² (Equation 2)
Step 3: Substitute a * t = 35 into Equation 2
From Equation 1, we know a * t = 35. Substitute this into Equation 2:
1500 = 20 * t + 0.5 * (a * t) * t
1500 = 20 * t + 0.5 * 35 * t
1500 = 20 * t + 17.5 * t
1500 = 37.5 * t
t = 1500 / 37.5 = 40 seconds
Step 4: Find Acceleration (a)
From Equation 1:
a * t = 35
Substitute t = 40:
a * 40 = 35
a = 35 / 40 = 0.875 m/s²
Final Answers:
- The train is in the tunnel for 40 seconds.
- The acceleration of the train is 0.875 m/s².
Verification:
To ensure the answers are correct, let’s verify the calculations.
1. Time Calculation:
– From Equation 1: a * t = 35.
– From Equation 2: 1500 = 20 * t + 17.5 * t.
– Solving for t: t = 40 seconds.
2. Acceleration Calculation:
– From Equation 1: a = 35 / 40 = 0.875 m/s².
3. Check Distance:
– Using Distance = Initial Velocity * Time + 0.5 * Acceleration * Time²:
– Distance = 20 * 40 + 0.5 * 0.875 * (40)².
– Distance = 800 + 0.5 * 0.875 * 1600.
– Distance = 800 + 700 = 1500 metres.
The calculations are consistent, and the answers are correct.
