A car that weighs 1.5 × 10^4 N is initially moving at a speed of 43 km/h when the brakes are applied and the car is brought to a stop in 16 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)
I wil be glad to critique your work.
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. We can use this equation to find the magnitude of the force that stops the car.
Step 1: Convert the given speed from km/h to m/s.
- Given speed = 43 km/h
- Conversion: 1 km/h = 1000 m/3600 s ≈ 0.2778 m/s
- Speed in m/s = 43 km/h × 0.2778 m/s = 11.94 m/s
Step 2: Calculate the magnitude of the force using Newton's second law of motion.
- Mass = Weight / Acceleration due to gravity
- Weight of the car = 1.5 × 10^4 N
- Acceleration due to gravity, g = 9.8 m/s^2 (assuming Earth's gravity)
- Mass = 1.5 × 10^4 N / 9.8 m/s^2 = 1530.61 kg
- Using Newton's second law: Force = Mass × Acceleration
- Force = 1530.61 kg × 11.94 m/s = 18317.32 N
(a) Therefore, the magnitude of the force that stops the car is approximately 18317.32 Newtons.
Step 3: Calculate the time required for the change in speed.
- Given stopping distance = 16 m
- Using the equation of motion: Distance = (Initial velocity × Time) + (0.5 × Acceleration × Time^2)
- Since the car is brought to a stop and the final velocity is zero, the equation becomes: Distance = Initial velocity × Time
- Rearranging the equation: Time = Distance / Initial velocity
- Time = 16 m / 11.94 m/s ≈ 1.34 seconds
(b) Therefore, the time required for the change in speed is approximately 1.34 seconds.
Step 4: Find the factors by which the stopping distance and stopping time are multiplied when the initial speed is doubled.
- Doubling the initial speed means the new initial speed is 2 times the initial speed mentioned above.
- New initial speed = 2 × 11.94 m/s = 23.88 m/s
- Similarly, the force applied is the same as before.
- New stopping distance = (New initial velocity × New time) = (23.88 m/s × Time)
- New stopping distance = 2 × (Initial stopping distance) = 2 × 16 m = 32 m
(c) The stopping distance is multiplied by a factor of 2 when the initial speed is doubled.
- New stopping time = New stopping distance / New initial velocity = (32 m) / (23.88 m/s) ≈ 1.34 seconds
(d) The stopping time remains the same when the initial speed is doubled.
So, the factors by which the stopping distance and stopping time are multiplied when the initial speed is doubled are:
(c) The stopping distance is multiplied by a factor of 2.
(d) The stopping time remains the same.
This problem highlights the danger of driving at high speeds, as doubling the initial speed doubles the stopping distance, leading to increased risk and less time for the driver to react.