ACTAR Mathematical & Physical Principles of Reconstruction 4 — Questions and Answers
Question 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)?
- 24.0 mph
- 36.0 mph (Correct answer)
- 40.5 mph
- 52.8 mph
Correct answer: 36.0 mph
COLM: m1v1 = (m1+m2)v2; v2 = (3600 × 88)/(3600+2400) = 316,800/6000 = 52.8 ft/s ≈ 36 mph.
Question 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:
- Wheel lockup direction
- Uphill adding to braking, downhill reducing braking effectiveness (Correct answer)
- Lateral offset of center of gravity
- Difference in front and rear axle friction
Correct 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).
Question 3: Impulse is defined as the product of:
- Mass and velocity
- Force and displacement
- Force and time (Correct answer)
- Acceleration and time
Correct answer: Force and time
Impulse J = F × Δt, and by the impulse-momentum theorem, impulse equals the change in momentum (J = Δp).
Question 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?
- SSF = T / (2 × hcg) (Correct answer)
- SSF = hcg / T
- SSF = 2 × hcg / T
- SSF = T² / hcg
Correct answer: SSF = T / (2 × hcg)
SSF = T/(2hcg); a higher SSF (wide track, low CG) indicates greater resistance to rollover.
Question 5: A vehicle in a 90° turn at constant speed experiences which type of acceleration?
- Tangential acceleration only
- No acceleration because speed is constant
- Centripetal (centrifugal) acceleration directed toward the center of curvature (Correct answer)
- Angular acceleration equal to g
Correct 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.
Question 6: The 'f' value (drag factor) on a surface is numerically equal to the coefficient of friction only when:
- The vehicle is on a level surface (Correct answer)
- All four tires are locked
- The vehicle's weight distribution is perfectly equal front-to-rear
- Temperature is above 60°F
Correct 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).
Question 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?
- 19.8 mph
- 29.4 mph (Correct answer)
- 31.7 mph
- 41.2 mph
Correct 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.
ACTAR Mathematical & Physical Principles of Reconstruction 4
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)?