Large decelerations produce large forces
Definition
Deceleration
Deceleration is acceleration that reduces the magnitude of an object's velocity.
- A rapid stop changes velocity in a short time. The magnitude of acceleration is ∣a∣=∣v−ut∣|a|=\left|\dfrac{v-u}{t}\right|∣a∣=tv−u, so a shorter stopping time produces a larger deceleration.
- A large deceleration requires a large resultant force. For an object of fixed mass, F=maF=maF=ma, so the force magnitude increases in direct proportion to the deceleration magnitude.
- Occupants continue moving when the vehicle first stops. Their inertia keeps them moving at the original velocity until a seat belt, airbag or part of the vehicle exerts a resultant force.
- Large forces can injure the body. Injury is more likely when the force is concentrated over a small area or when different body tissues stop over different times.
Safety features reduce the force
Definition
Stopping time
Stopping time is the time taken for an object or person to change from its initial velocity to its final velocity during a stop.
- Seat belts stretch slightly. This increases the time and distance over which the occupant loses momentum, reducing the average force because F=ΔpΔtF=\dfrac{\Delta p}{\Delta t}F=ΔtΔp.
- Airbags compress and spread the force. Compression increases the stopping time, while the large contact area reduces pressure on the head and chest because p=FAp=\dfrac{F}{A}p=AF.
- Crumple zones deform. Energy is transferred by work done deforming the vehicle, and the longer collision time reduces the force on the passenger compartment.
- A rigid passenger cell limits deformation around occupants. It reduces the chance that the vehicle structure enters the space occupied by passengers.
Example
Estimating a collision force
- Situation: a 1200 kg1200\ \text{kg}1200 kg car slows from 13 m s−113\ \text{m s}^{-1}13 m s−1 to rest in 0.20 s0.20\ \text{s}0.20 s.
- Acceleration: a=0−130.20=−65 m s−2a=\dfrac{0-13}{0.20}=-65\ \text{m s}^{-2}a=0.200−13=−65 m s−2.
- Resultant force: F=ma=1200×(−65)=−7.8×104 NF=ma=1200\times(-65)=-7.8\times10^{4}\ \text{N}F=ma=1200×(−65)=−7.8×104 N.
- Interpretation: the negative sign shows that the force acts opposite to the initial motion, and its magnitude is 7.8×104 N7.8\times10^{4}\ \text{N}7.8×104 N.
Exam technique
Explaining the danger
- Link the quantities in order: short stopping time, large change in velocity per second, large deceleration, large resultant force, greater risk of injury.
- Name the object experiencing the force. A correct answer distinguishes the force on the vehicle from the force on an occupant.
- When estimating, state sensible values and show the method. Credit can be earned for a valid calculation even if a reasonable estimate differs from another estimate.
Common Mistake
- Do not say that a safety feature reduces the change in momentum. A person brought to rest has the same momentum change, but the change occurs over a longer time.
- Do not use speed alone to explain force. Force depends on mass and acceleration, while acceleration depends on the change in velocity and the time taken.
- Do not confuse force with pressure. An airbag can reduce both by increasing stopping time and contact area, but these are separate effects.
Self review
- How does a shorter stopping time affect deceleration for the same change in velocity?
- Why does a larger deceleration produce a larger resultant force for the same mass?
- How does a seat belt reduce the average force on an occupant?
- Why does an airbag reduce pressure as well as force?
