15. A car starting from rest accelerates at the rate R through a distance S, then continues at a
constant speed for time t and then decelerates at rate R/2 to come to rest. If the total distance
travelled is 15S, then:
a.S= 1/72 Rt2
b.S= 1/12 Rt2
c.S= 1/36 Rt2
d.None of above
Answer: a. S = 1/72 Rt²
A car accelerating from rest at rate R through distance S reaches speed v₁ = √(2RS), travels distance d₂ = v₁t at constant speed for time t, then decelerates at R/2 covering d₃ = v₁²/R, with total distance 15S yielding the solution .
A car accelerating from rest at rate R through distance S reaches speed v₁ = √(2RS), travels distance d₂ = v₁t at constant speed for time t, then decelerates at R/2 covering d₃ = v₁²/R, with total distance 15S yielding the solution .
Step-by-Step Kinematics Solution
Phase 1 (Acceleration): v₁² = 2RS, so v₁ = √(2RS)
Phase 2 (Constant speed): Distance d₂ = v₁t = √(2RS)·t
Phase 3 (Deceleration): Final velocity 0, acceleration -R/2 gives 0 = v₁² – R·d₃, so d₃ = v₁²/R = 2S
Total distance equation: S + √(2RS)·t + 2S = 15S simplifies to √(2RS)·t + 3S = 15S .
Solving: √(2RS)·t = 12S, square both sides: 2RSt² = 144S², so S = Rt²/72 .
Option Analysis
Options test algebraic manipulation of the distance equation.
- a. S = 1/72 Rt²: Matches exact solution from √(2RS)t = 12S after squaring and simplifying
- b. S = 1/12 Rt²: Overestimates by factor of 6; likely from incorrect deceleration distance calculation
- c. S = 1/36 Rt²: Wrong coefficient; possibly confusing deceleration rate R/2 effect
- d. None of above: Incorrect since option a derives precisely from physics
Distance Breakdown Table
| Phase | Distance Formula | Expression | Contribution |
|---|---|---|---|
| 1: Acceleration | S | S | 1S |
| 2: Constant speed | v₁t | √(2RS)t | ~11S |
| 3: Deceleration | v₁²/R | 2S | 2S |
| Total | 15S | S = Rt²/72 | 15S |
Final Answer: a. S = 1/72 Rt²
