There are two factors that contribute to the total stopping distance for a traveling vehicle. These two factors are the perception-reaction distance and the braking distance. When an event occurs that requires an emergency stop, the vehicle continues to travel at its initial velocity while the driver reacts to the event. The distance traveled for this time is the perception-reaction distance. Secondly, the vehicle undergoes constant acceleration while the brakes are applied. The distance traveled for this time interval is the braking distance. Historically, engineers have used a perception-reaction time of 0.75 seconds, but they now assume a perception-reaction time of 1.0 seconds for the average driver. Write an expression for the total stopping distance of the vehicle.
The Correct Answer and Explanation is:
Correct Answer:
Let:
- vvv be the initial velocity of the vehicle (in meters per second),
- ttt be the perception-reaction time (usually 1.0 second),
- aaa be the constant deceleration due to braking (in meters per second squared).
The total stopping distance ddd is the sum of:
- Perception-reaction distance: d1=v⋅td_1 = v \cdot td1=v⋅t
- Braking distance: d2=v22ad_2 = \frac{v^2}{2a}d2=2av2
So, the expression for total stopping distance is:d=v⋅t+v22ad = v \cdot t + \frac{v^2}{2a}d=v⋅t+2av2
Explanation
When a driver notices an event that requires stopping, the car does not stop instantly. The vehicle first travels during the driver’s reaction time, then slows down to a complete stop due to braking. This entire process contributes to what is called the total stopping distance.
The first part of this distance is the perception-reaction distance, which depends only on the speed of the vehicle and how long it takes the driver to respond. Engineers currently assume this response time to be 1.0 second for an average driver. During this reaction time, the vehicle continues to move at its original speed. If the car is moving at velocity vvv, and the reaction time is ttt, then the car travels a distance d1=v⋅td_1 = v \cdot td1=v⋅t.
Once the brakes are applied, the vehicle begins to decelerate. This deceleration is assumed to be constant and can be represented as aaa. The distance the car travels while slowing down is called the braking distance, and it can be calculated using the physics formula for motion: d2=v22ad_2 = \frac{v^2}{2a}d2=2av2. This comes from rearranging the equation v2=2adv^2 = 2adv2=2ad used when an object slows to rest under constant acceleration.
The total stopping distance is the sum of both distances: perception-reaction and braking. Therefore, the full expression becomes:Total stopping distance=v⋅t+v22a\text{Total stopping distance} = v \cdot t + \frac{v^2}{2a}Total stopping distance=v⋅t+2av2
This formula is essential for designing safe roads, setting speed limits, and understanding how long it takes a vehicle to stop in emergencies.
