You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was traveling eastward. Car B weighs 1125 lb and was traveling westward at 45.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 22.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. How fast (in miles per hour) was car A traveling just before the collision? (This problem uses English units because they would be used in a U.S. legal proceeding.)
Well, it sounds like Car A and Car B really locked horns, or rather, bumpers in this case! Let's get to the bottom of this collision puzzle, shall we?
To find the initial speed of Car A, we need to use some good old physics. First, let's convert the distance traveled by the cars into miles. Since there are 5,280 feet in a mile, 22.5 ft would be 22.5/5280 miles. Let me just pull out my handy-dandy calculator... Ah, yes! Approximately 0.00426 miles.
Now that we have the distance, let's figure out the deceleration of the cars. The net force causing this deceleration is the force of friction. Using the formula F = μmg, where μ is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity, we can find the deceleration.
For Car A: F = μ(mass of Car A)(gravity)
For Car B: F = μ(mass of Car B)(gravity)
Now let's substitute in the values:
For Car A: F_A = (0.750)(1515 lb)(32 ft/s^2)
For Car B: F_B = (0.750)(1125 lb)(32 ft/s^2)
Now, we know that the net force experienced by both cars is the same since they're locked together. So, F_A = F_B. We can equate the two expressions for force and solve for the mass of Car A:
(0.750)(1515 lb)(32 ft/s^2) = (0.750)(1125 lb)(32 ft/s^2)
Now, I'm about to unleash my math skills to solve this, so hold on tight! After some mathematical magic, let me reveal the mass of Car A... Drumroll, please! Car A weighs approximately 1125 lb.
Now comes the fun part! We will use physics equations again to find the initial velocity of Car A. The equation looks like this:
Vf^2 = Vi^2 + 2ad
Where Vf is final velocity (0 mph since the cars come to a stop), Vi is initial velocity, a is acceleration (deceleration in our case), and d is the distance traveled.
Now let's plug in the values we know:
0 = Vi^2 + 2(-32 ft/s^2)(0.00426 miles)
By solving this equation for Vi, we will arrive at the initial velocity of Car A just before the collision. After doing some calculations, Vi comes out to be approximately -15.4 mph. Oh dear, Car A was going in reverse before the collision!
To find the actual speed, we just take the magnitude of -15.4 mph, which is 15.4 mph. So, Car A was traveling at approximately 15.4 mph just before the collision.
But remember, my dear friend, that this is all a fictional scenario, and real-life situations can have many variables. So, let's keep the clownish physics in the circus and leave the serious stuff for the courtroom!
To determine the speed of car A just before the collision, we can use the concept of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
1. We need to find the initial velocity of car A before the collision. Let's assume it to be v mph.
2. The momentum of car A before the collision is given by the formula: momentum = mass × velocity.
The momentum of car A before the collision = (1515 lb) × (v mph).
3. Car B is traveling at a speed of 45.0 mph westward. As the cars locked bumpers and slid eastward, their effective velocity will be the difference between their individual velocities.
The effective velocity after the collision = (45.0 mph) - (v mph) = (45.0 - v) mph.
The momentum of car B before the collision is given by the formula: momentum = mass × velocity.
The momentum of car B before the collision = (1125 lb) × (45.0 mph).
4. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
(1515 lb) × (v mph) = (1125 lb) × (45.0 - v) mph.
5. Let's solve this equation to find the value of v, which represents the velocity of car A just before the collision.
(1515 lb) × (v mph) = (1125 lb) × (45.0 - v) mph.
We need to convert the units to be consistent. Since the mass is in pounds, we should convert the velocity to feet per second (fps) for consistent units since 1 mph = 1.47 fps.
(1515 lb) × (v × 1.47 fps) = (1125 lb) × [(45.0 mph) - (v × 1.47 fps)].
6. Now, let's solve for v by simplifying the equation:
2224.05 v = 50287.5 - 1653.75 v
2224.05 v + 1653.75 v = 50287.5
3877.8 v = 50287.5
v = 50287.5 / 3877.8
v ≈ 12.9709
7. Therefore, car A was traveling at approximately 12.9709 mph just before the collision.
To determine the speed of car A just before the collision, we can use the principles of physics and the given information.
First, let's convert the weight of car A and car B from pounds to slugs. We use the conversion factor 1 slug = 32.2 lb:
Weight of car A (W_A) = 1515 lb / 32.2 lb/slug = 47.04 slugs
Weight of car B (W_B) = 1125 lb / 32.2 lb/slug = 34.79 slugs
Next, let's convert the distance traveled by the cars from feet to miles. We use the conversion factor 1 mile = 5280 ft:
Distance traveled (d) = 22.5 ft / 5280 ft/mile = 0.00426 miles
Now, let's calculate the frictional force acting on the cars. The frictional force (F_friction) can be calculated using the equation:
F_friction = coefficient of kinetic friction (μ_k) * Normal force
The normal force (N) can be calculated as the weight of car B multiplied by the cosine of the angle between the weight vector and the horizontal plane. Since the cars collide head-on, the angle is 180 degrees, and the cosine of 180 degrees is -1. Therefore:
N = W_B * (-1) = -34.79 slugs
Substituting the values into the equation, we have:
F_friction = 0.750 * -34.79 slugs = -26.09 slugs
Since the frictional force acts in the opposite direction of car A's motion, it will provide a negative acceleration. Using Newton's second law of motion, we can calculate the acceleration (a) as:
F_friction = ma
-26.09 slugs = 47.04 slugs * a
a = -0.554 ft/s^2
Finally, we can find the initial velocity of car A (v_Ai) using the equation of motion:
v_Af^2 = v_Ai^2 + 2ad
Since the final velocity (v_Af) is zero (cars come to a stop), the equation becomes:
0 = v_Ai^2 + 2 (-0.554 ft/s^2) (0.00426 miles * 5280 ft/mile)
0 = v_Ai^2 - 24.45 ft^2/s^2
Solving for v_Ai, we find:
v_Ai = 4.944 ft/s
Finally, let's convert the velocity from feet per second to miles per hour. We use the conversion factor 1 mile = 5280 ft and 1 hour = 3600 seconds:
v_Ai = 4.944 ft/s * (1 mile/5280 ft) * (3600 s/1 hour) ≈ 3.37 mph
Therefore, car A was traveling at approximately 3.37 miles per hour just before the collision.
east is positive
1515 E -1125(45) = 2640 V
F = 2640*.75 = -1980 pounds
so
-1980 pounds = m a
but m = 2640/g where g = 32.2 ft/s^2
so
a = -(1980/2640)g = -.75 g = -24.15 ft/s^2
= dv/dt
0 = V-24.15 t so V =24.15 t
22.5 = V t -12.07 t^2
22.5 = 24.15 t^2 - 12.07 t^2 = 12.07 t^2
t = 1.37 seconds
so
V = 24.15 * 1.37 seconds = 33.1 ft/s
convert that to mph
33.1 ft/s = 22.6 mph
now back to momentum
1515 E -1125(45) = 2640 V
1515 E = 50625+59664
E = 72.8 mph