The Problem:
A cyclist is free-wheeling down a long straight hill. The times between passing successive kilometre posts are 100 seconds and 80 seconds. Assuming his acceleration is constant, find this acceleration.
Douglas Quadling Mechanics1 Exercise1D Q8
A cyclist is free-wheeling down a long straight hill. The times between passing successive kilometre posts are 100 seconds and 80 seconds.
Problem Statement:
A cyclist is free-wheeling down a long straight hill. The times between passing successive kilometre posts are 100 seconds and 80 seconds. Assuming his acceleration is constant, find this acceleration.
Solution:
Step 1: Equation for the First Kilometre
The distance covered in the first kilometre is 1000 meters. Using the equation of motion:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time²
For the first kilometre:
1000 = u * 100 + 0.5 * a * (100)²
1000 = 100u + 5000a (Equation 1)
Step 2: Velocity at the End of the First Kilometre
The velocity at the end of the first kilometre (v₁) is:
v₁ = u + a * t₁
v₁ = u + 100a
Step 3: Equation for the Second Kilometre
For the second kilometre, the distance is also 1000 meters, and the time taken is 80 seconds. Using the equation of motion:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time²
Substitute v₁ = u + 100a and t = 80:
1000 = (u + 100a) * 80 + 0.5 * a * (80)²
1000 = 80u + 8000a + 3200a
1000 = 80u + 11200a (Equation 2)
Step 4: Solve the System of Equations
We now have two equations:
1. 1000 = 100u + 5000a (Equation 1)
2. 1000 = 80u + 11200a (Equation 2)
From Equation 1:
100u = 1000 – 5000a
u = 10 – 50a (Equation 3)
Substitute Equation 3 into Equation 2:
1000 = 80(10 – 50a) + 11200a
1000 = 800 – 4000a + 11200a
1000 = 800 + 7200a
200 = 7200a
a = 200 / 7200 = 1/36 m/s²
Final Answer:
The cyclist’s acceleration is 1/36 m/s².
Verification:
To ensure the answer is correct, let’s verify the calculations.
1. From Equation 3:
u = 10 – 50a = 10 – 50 * (1/36) = 310/36 m/s
2. Check Equation 1:
100u + 5000a = 100 * (310/36) + 5000 * (1/36) = 36000/36 = 1000 m
3. Check Equation 2:
80u + 11200a = 80 * (310/36) + 11200 * (1/36) = 36000/36 = 1000 m
Both equations are satisfied, confirming that the acceleration is correct.
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