1D Kinematics: Solving Uniform Acceleration Problems
1D Kinematics: Acceleration and Braking
This simulation models a car accelerating from rest at 2 m/s² until it reaches 20 m/s, at which point it immediately brakes at -1 m/s² until it comes to a complete stop.
Physics Insights
Press Start to observe the motion. Pay attention to how the graphs change shape during each phase!
Kinematics of a Car in 1D
Q. Suppose the car in Problem 2 is traveling at twice the initial speed (49.2 m/s) but applies the brakes with the same constant acceleration. By what factor does the stopping distance change? Justify your answer without calculating the exact new distance.
Using the kinematic equation v² = v₀² + 2aΔx and setting the final velocity v = 0, we can rearrange the equation to solve for stopping distance: Δx = -v₀² / (2a). Because the 'a' remains constant, the stopping distance is directly proportional to the square of the initial velocity
Δx ∝ v₀²
If the initial speed is doubled (increased by a factor of 2), the stopping distance will increase by a factor of 2², which is 4. The new stopping distance is four times greater.
Q. A car starts from rest and accelerates at a rate +a₁ until it reaches a maximum speed v_max. It immediately brakes with an acceleration of -a₂ until it stops. Derive an algebraic expression for the total distance D traveled by the car in terms of a₁, a₂, and v_max.
For the first phase of motion (accelerating from rest to v_max), we use v² = v₀² + 2aΔx. Substituting our variables gives v_max² = 0² + 2(a₁)d₁. Solving for d₁ yields
d₁ = v_max² / (2a₁).
For the second phase (braking from v_max to rest), the equation becomes 0² = v_max² + 2(-a₂)d₂. Solving for d₂ yields
d₂ = v_max² / (2a₂).
The total distance D is the sum of the distances from both phases:
D = d₁ + d₂ = (v_max² / 2) × (1/a₁ + 1/a₂)
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