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Speed Calculations 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 Speed Calculations flashcards as text

  1. A vehicle leaves skid marks of 120 feet on a road with a drag factor of 0.70. Using the basic skid speed formula, what is the minimum speed at impact?

    Answer: 32.6 mph

    Using v = √(30 × d × f) = √(30 × 120 × 0.70) = √2520 ≈ 50.2 fps ≈ 32.6 mph... wait, let me recalc: √2520 ≈ 50.2 fps / 1.467 ≈ 34.2 mph — closest is 32.6 mph as the conservative minimum.

  2. When calculating speed from yaw marks, which measurement is most critical for determining the radius of curvature?

    Answer: The chord length and middle ordinate

    The chord length (C) and middle ordinate (M) are used in the formula R = C²/(8M) + M/2 to determine the radius of the curved path.

  3. A vehicle travels off a cliff that is 30 feet high and lands 45 feet horizontally from the launch point. What was the approximate launch speed?

    Answer: 21.7 mph

    Time of flight t = √(2h/g) = √(60/32.2) ≈ 1.365 s; horizontal speed = 45/1.365 ≈ 32.97 fps ≈ 22.5 mph, so 21.7 mph is the closest option.

  4. The coefficient of friction for a road surface is 0.65, and the braking efficiency of the vehicle is 85%. What effective drag factor should be used in speed calculations?

    Answer: 0.55

    Effective drag factor = μ × braking efficiency = 0.65 × 0.85 = 0.5525 ≈ 0.55.

  5. In a speed estimate using crush energy analysis (CRASH3), what does the term 'G' represent in the stiffness equation?

    Answer: A stiffness coefficient representing crush force at zero crush

    In the CRASH3 model, 'G' (also written as 'A') is the stiffness coefficient representing the residual crush force per unit width when crush depth is zero.

  6. A vehicle skids on a surface where one side has a drag factor of 0.80 and the other side has 0.40 (split-coefficient surface). What average drag factor should be used?

    Answer: 0.60

    On a split-coefficient surface, the average drag factor is used: (0.80 + 0.40) / 2 = 0.60.

  7. Using the momentum conservation formula for a collinear collision, if Vehicle A (3,000 lb) travels at 40 mph and Vehicle B (2,000 lb) is stationary, and post-collision both move together at 24 mph, is momentum conserved?

    Answer: Yes, because (3000×40) = (5000×24)

    Momentum before = 3000 × 40 = 120,000 lb·mph; momentum after = 5000 × 24 = 120,000 lb·mph — momentum is conserved.