0
What is one factor the changes both the first part in the second part of the stopping distance of a car?
One key factor that affects both the first part (reaction distance) and the second part (braking distance) of a car's stopping distance is speed. As a vehicle's speed increases, the time it takes for a driver to react and the distance required to stop both increase significantly. This is due to the fact that higher speeds result in longer distances traveled during the driver's reaction time and greater kinetic energy that must be overcome during braking. Therefore, speed plays a critical role in determining the overall stopping distance of a vehicle.
What else can I help you with?
What Speeding is a factor in of all fatal accidents?
Speeding is a factor in all fatal accidents for multiple reasons. The main being the greater the speed, the greater force involved. More force increases the chance of great bodily harm or death in a collision. A second factor that speed changes is breaking distance. The faster a vehicle is moving, the longer it takes to stop. A vehicle traveling 30 MPH has an approximate stopping distance of 109 feet, 60 MPH 304 FT, and 90 MPH 584 FT.
What is the orbital distance of Mars in a second?
. . 'in a second' . . that bit makes the question impossible to answer. The distance changes with time but the orbit is stable.
On the expressway, maintain a 3-4 second following distance If the road is slippery or wet, adjust your distance to?
6-8 seconds
At 40 mph what is your vehicle's reaction distance?
At 40 mph, a vehicle's reaction distance, which is the distance covered from the moment a driver perceives a hazard to the moment they begin to brake, is approximately 44 feet. This calculation assumes a reaction time of about 1.5 seconds. Therefore, at this speed, a driver travels about 58 feet per second, leading to the reaction distance being a key factor in overall stopping distance.
What will the stopping distance be for a 3000kg car if 3000 N of force are applied when the car is traveling 10 ms?
The stopping distance for a 3000kg car if 3000 N of force is applied when the car is traveling 10 ms is 50 meter. This is based on Newton's second law of force.
How would the distance the car travels each second change if it were slowing down?
In that case, each second the distance travelled will be less than the second before that.
The stopping distance of a car when it is going at 50 miles per hour?
5o miles per hour is 73.3 feet per second. The average stopping distance for the breaks at that speed is 128 feet. Now add to that the average reaction time for a driver at that speed which is 3/4 second so we add 55 feet gives us a total of 183 feet.
What will the stopping distance be for 3000-kg car if -3000 N of force are applied when the car is traveling 10ms?
To calculate stopping distance, you need to know the deceleration of the car. Here, deceleration can be calculated using Newton's second law: deceleration = force / mass. With the given force of -3000 N and mass of 3000 kg, the deceleration would be -1 m/s^2. Using the equation of motion, final velocity^2 = initial velocity^2 + 2 * acceleration * distance, you can calculate the stopping distance.
At 40 mph what is your vehicle and reaction distance?
At 40 mph, a vehicle travels approximately 58 feet per second. The average reaction time for a driver is about 1.5 seconds, which means the reaction distance would be around 87 feet (1.5 seconds x 58 feet/second). Additionally, the stopping distance will vary depending on road conditions and vehicle braking capabilities. Therefore, at 40 mph, the total stopping distance can be around 120-140 feet when factoring in both reaction and braking distances.
What will the stopping distance for 1000-kg car moving 20 ms when you apply a force of -500 N?
Given a force of -500 N, which implies braking, the stopping distance of the car can be calculated using the equation ( d = v^2 / 2a ), where ( d ) is the stopping distance, ( v ) is the initial velocity (20 m/s), and ( a ) is the acceleration produced by the force. Using Newton's second law, we have ( a = F / m = -500 / 1000 = -0.5 , \text{m/s}^2 ). Substituting ( v = 20 , \text{m/s} ) and ( a = -0.5 , \text{m/s}^2 ) into the stopping distance equation, we get ( d = 20^2 / (2 \times 0.5) = 400 , \text{m} ). Hence, the stopping distance for the car will be 400 meters.
With each environmental factor you should do what following distance by second?
It is recommended to maintain a following distance of at least two seconds between your vehicle and the one in front of you in normal driving conditions. This allows for enough time to react in case of sudden braking or changes in traffic. Adjust this following distance based on weather conditions, traffic density, and road surface conditions.
You should increase or decrease with each additional environmental factor your following distance with 1 second?
Decrease
