Stopping Distances
In this lesson we learn about stopping distance and its components.
IN THIS LESSON:
Stopping distance
What it is
Where do we apply it
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Practice Questions
Question 1: Define the term 'stopping distance'.
Answer: Stopping distance is the total distance a vehicle travels from the moment the driver realises they need to stop to the moment the vehicle comes to a complete stop.
Question 2: What two distances make up the total stopping distance?
Answer: The total stopping distance is the sum of the thinking distance and the braking distance.
Question 3: Define 'thinking distance'.
Answer: Thinking distance is the distance a vehicle travels during the driver's reaction time – the time it takes them to notice the hazard and apply the brakes.
Question 4: Define 'braking distance'.
Answer: Braking distance is the distance a vehicle travels from the moment the brakes are applied to the moment it comes to a complete stop.
Question 5: A car is traveling at a constant velocity. Draw a velocity-time graph to represent the car's motion before the brakes are applied.
Answer: The velocity-time graph should show a horizontal line (constant velocity) above the x-axis.
Question 6: On the velocity-time graph from Question 5, sketch a line to show the car's motion after the brakes are applied, assuming constant deceleration.
Answer: The line should start from the end of the horizontal line and slope downwards at a constant rate (straight line) to the x-axis (where velocity = 0).
Question 7: How can you determine the braking distance from a velocity-time graph where the deceleration is constant?
Answer: The braking distance is equal to the area under the velocity-time graph from the point where the brakes are applied to the point where the velocity reaches zero.
Question 8: A car travels at 20 m/s. The driver's reaction time is 0.8 s. Calculate the thinking distance.
Answer Walkthrough:
- Given: Velocity (v) = 20 m/s, Reaction time (t) = 0.8 s.
- Formula: Distance = Speed x Time
- Calculation: Thinking distance = 20 m/s × 0.8 s = 16 m.
Answer: The thinking distance is 16 m.
Question 9: A car decelerates at a constant rate of 5 m/s2. Its initial velocity is 25 m/s. Calculate the braking distance.
Answer Walkthrough:
We can use the following SUVAT equation: v2=u2+2as, where:
- v = final velocity (0 m/s).
- u = initial velocity (25 m/s).
- a = acceleration (deceleration = −5 m/s2).
- s = braking distance (what we want to find).
Rearrange the formula: $s=(v^2-u^2)/(2a)$.
Calculation: $s=(0^2-25^2)/(2 \times -5)=(-625)/(-10)=62.5 \text{ m}$.
Answer: The braking distance is 62.5 m.
Question 10: A car has a thinking distance of 12 m and a braking distance of 45 m. What is the total stopping distance?
Answer Walkthrough:
- Formula: Stopping distance = Thinking distance + Braking distance.
- Calculation: Stopping distance = 12 m + 45 m = 57 m.
Answer: The total stopping distance is 57 m.
Question 11: A driver sees a hazard and applies the brakes after 1.1 s. The car was traveling at 18 m/s and decelerates at 4 m/s2. Calculate the total stopping distance.
Answer Walkthrough:
Thinking Distance:
- Formula: Distance = Speed x Time
- Calculation: Thinking distance = 18 m/s × 1.1 s = 19.8 m.
Braking Distance:
- $v^2=u^2+2as$
- $s=(v^2-u^2)/(2a)$
- $s=(0^2-18^2)/(2 \times -4)=(-324)/(-8)=40.5 \text{ m}$.
Total Stopping Distance:
- Stopping distance = Thinking distance + Braking distance.
- Stopping distance = 19.8 m + 40.5 m = 60.3 m.
Answer: The total stopping distance is 60.3 m.
Question 12: A car travels at 30 m/s. The braking distance is 75 m. Assuming constant deceleration, calculate the deceleration of the car.
Answer Walkthrough:
- $v^2=u^2+2as$
- $a=(v^2-u^2)/(2s)$
- $a=(0^2-30^2)/(2 \times 75)=(-900)/(150)=-6 \text{ m/s}^2$.
Answer: The deceleration is $6 \text{ m/s}^2$.
Question 13: A velocity-time graph shows a car decelerating from 22 m/s to 0 m/s in 5.5 s. Calculate the braking distance.
Answer Walkthrough:
The area under the graph is a triangle.
- Formula: Area of triangle = 0.5 x base x height.
- Calculation: Braking distance = 0.5 × 5.5 s × 22 m/s = 60.5 m.
Answer: The braking distance is 60.5 m.
Question 14: A lorry has a thinking distance of 20 m and a braking distance that is three times its thinking distance. Calculate the total stopping distance.
Answer Walkthrough:
- Braking Distance: 20 m × 3 = 60 m.
- Total Stopping Distance: 20 m + 60 m = 80 m.
Answer: The total stopping distance is 80 m.
Question 15: A car traveling at 15 m/s has a braking distance of 30 m. If the car doubles its speed to 30 m/s, what will be the new braking distance, assuming the same deceleration?
Answer Walkthrough:
- We first find the deceleration at 15 m/s: $a=(v^2-u^2)/(2s)=(0^2-15^2)/(2 \times 30)=-225/60=-3.75 \text{ m/s}^2$.
- Now, we calculate the braking distance at 30 m/s with the same deceleration: $s=(v^2-u^2)/(2a)=(0^2-30^2)/(2 \times -3.75)=-900/-7.5=120 \text{ m}$.
Answer: The new braking distance will be 120 m.