The Problem:

A particle is moving along a straight line with constant acceleration. In an interval of T seconds it moves D metres; in the next interval of 3T seconds it moves 9D metres. How far does it move in a further interval of T seconds?

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Douglas Quadling Mechanics 1
Exercise 1D Q10

A particle is moving along a straight line with constant acceleration. In an interval of T seconds it moves D metres; in the next interval of 3T seconds it moves 9D metres…

Problem Statement:
A particle is moving along a straight line with constant acceleration. In an interval of T seconds, it moves D metres; in the next interval of 3T seconds, it moves 9D metres. How far does it move in a further interval of T seconds?

Solution:

Step 1: First Interval (T seconds, distance D)
Using the equation of motion:

Distance = Initial Velocity * Time + 0.5 * Acceleration * Time²

For the first interval:
D = u * T + 0.5 * a * T²   (Equation 1)

The velocity at the end of the first interval (v₁) is:
v₁ = u + a * T   (Equation 2)

Step 2: Second Interval (3T seconds, distance 9D)
Using the equation of motion for the second interval:

9D = v₁ * 3T + 0.5 * a * (3T)²

Substitute v₁ = u + a * T:
9D = (u + a * T) * 3T + 0.5 * a * 9T²
9D = 3uT + 3aT² + 4.5aT²
9D = 3uT + 7.5aT²   (Equation 3)

Step 3: Solve Equations 1 and 3
From Equation 1:
D = uT + 0.5aT²

Multiply Equation 1 by 3:
3D = 3uT + 1.5aT²   (Equation 4)

Subtract Equation 4 from Equation 3:
9D – 3D = (3uT + 7.5aT²) – (3uT + 1.5aT²)
6D = 6aT²
a = D / T²

Step 4: Find Initial Velocity (u)
Substitute a = D / T² into Equation 1:
D = uT + 0.5 * (D / T²) * T²
D = uT + 0.5D
uT = 0.5D
u = D / (2T)

Step 5: Velocity at the End of the Second Interval (v₂)
Using Equation 2 and the value of a:
v₁ = u + aT = D / (2T) + (D / T²) * T = D / (2T) + D / T = 3D / (2T)

The velocity at the end of the second interval (v₂) is:
v₂ = v₁ + a * 3T = 3D / (2T) + (D / T²) * 3T = 3D / (2T) + 3D / T = 9D / (2T)

Step 6: Distance in the Further Interval of T Seconds
Using the equation of motion for the further interval:
Distance = v₂ * T + 0.5 * a * T²

Substitute v₂ = 9D / (2T) and a = D / T²:
Distance = (9D / (2T)) * T + 0.5 * (D / T²) * T²
Distance = 9D / 2 + D / 2 = 10D / 2 = 5D

Final Answer:
The particle moves 5D metres in the further interval of T seconds.

Verification:
To ensure the answer is correct, let’s verify the calculations.

1. From Step 4, u = D / (2T) and a = D / T².
2. From Step 5, v₂ = 9D / (2T).
3. From Step 6, the distance in the further interval is 5D.

All steps are consistent, and the calculations are correct. The final answer is 5D.