Practice Test Geeks(ACTAR) Accreditation Commission for Traffic Accident Reconstructionist Practice Test
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Mathematical & Physical Principles of Reconstruction Flashcards

7 cards from real ACTAR practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Mathematical & Physical Principles of Reconstruction flashcards as text

  1. A 3,600-lb vehicle traveling at 60 mph collides with a stationary 2,400-lb vehicle, and they move together post-impact. What is the combined post-impact speed (use 88 ft/s for 60 mph)?

    Answer: 36.0 mph

    COLM: m1v1 = (m1+m2)v2; v2 = (3600 × 88)/(3600+2400) = 316,800/6000 = 52.8 ft/s ≈ 36 mph.

  2. The equation for stopping distance on a grade where θ is the road angle is S = v²/[2g(f ± sin θ)]. The ± accounts for:

    Answer: Uphill adding to braking, downhill reducing braking effectiveness

    On a downgrade, gravity assists motion and reduces effective braking (minus); on an upgrade, gravity opposes motion and aids braking (plus).

  3. Impulse is defined as the product of:

    Answer: Force and time

    Impulse J = F × Δt, and by the impulse-momentum theorem, impulse equals the change in momentum (J = Δp).

  4. In a rollover event, the vehicle's center of gravity height (hcg) and track width (T) determine the static stability factor (SSF). Which formula is correct?

    Answer: SSF = T / (2 × hcg)

    SSF = T/(2hcg); a higher SSF (wide track, low CG) indicates greater resistance to rollover.

  5. A vehicle in a 90° turn at constant speed experiences which type of acceleration?

    Answer: Centripetal (centrifugal) acceleration directed toward the center of curvature

    Even at constant speed, circular motion requires centripetal acceleration directed toward the center of the curve to continuously change the velocity direction.

  6. The 'f' value (drag factor) on a surface is numerically equal to the coefficient of friction only when:

    Answer: The vehicle is on a level surface

    On a level surface, the normal force equals the vehicle weight, so drag factor = friction force / weight = μ (coefficient of friction).

  7. A vehicle slides sideways 40 feet with a drag factor of 0.72 after impact. Using the skid formula, what was the approximate lateral speed at that point?

    Answer: 29.4 mph

    S = √(30 × 40 × 0.72) = √864 ≈ 29.4 mph; the skid formula applies equally to lateral sliding with the appropriate drag factor.