In this accident, a Porsche Carrera approached an intersection from a minor road
ID: 1968791 • Letter: I
Question
In this accident, a Porsche Carrera approached an intersection from a minor road while a Saturn SL approached the same intersection from a major road, traveling south 70 foot long skid marks exist going off at a 70 degree angle from the collision point, as shown in figure 1. The skids indicate that the two vehicles were locked together from the moment of impact until coming to rest. It is a dry, sunny day and the road is composed of asphalt. Based on this information, your fellow officer has already calculated that the force of friction between the road surface and the tires can provide a maximal acceleration or deceleration of 6.4 m/sec2. The car on the minor road (Porsche) has a stop sign at the intersection and the major road has a speed limit of 45 mph. The cars were presumably travelling straight just before the collision. The collision occurred towards the left-hand side of the major road roughly one quarter of the way across. The two cars have essentially the same mass.According to the manufacturer’s specifications, a Porsche Carrera can accelerate from 0 to 60 mph in 5.3 seconds.a)What is the rate of acceleration for the Porsche in m/s2
b)What is the skid mark distance in meters?
c)What is the velocity of the combined mass immediately after the collision?
d)What is the velocity of the Porsche just before the collision?
e)What is the distance that the Porsche needed to accelerate to that speed?
f)What was the speed of the Saturn?
Explanation / Answer
This is the wordiest problem I've read in a long time. The bets I could do on short time was this, hope it helps. a.) first we need unit conversions. 1 mile= 1609344 meters 1hr= 3600 seconds. Second acceleration is the change in velocity (60-0) over time (5.3-0) soo... (1609344 meters) * (60 miles) divided by (3600 seconds)*(5.3 seconds) equals: acceleration of 5.06 m/s^2 b.) Unit conversions: 1 foot= 0.3048 meters. (0.3048 meters)*(70ft) equals 21.34 meters c.) Distance of collision can be solved using a constant acceleration kinematic equation. v^2= v(initial)^2+2a*(s2-s1) solve for initial velocity. final velocity is zero. delta s is 21.34 meters and accel is -6.4 m/s^2 V^2(initial) + 2*(-6.4)*(21.34)=0 V^2(initial)= 273.152 V(initial)= sqrt(273.152) V(initial)= 16.53 meters/second Note that this V initial is the initial velocity AFTER the collision. d.) To solve the rest of this it seems there is some missing information. Maybe from the diagram? It would be helpful to know what the distance of 1/4 through the intersection is. For this, it is common in physics to use right angle geometry to solve. However, this might help you with the rest of the problem if you are able to find the porche's velocity at the collision. After you find one cars velocity at the collision you can use conservation of momentum to solve. m*v1 +m*v2= m*v' since the mass of the two vehicles are the same and since this is an elastic collision (both cars stick together the final mass is m1+m2. Since m is the same we can right 2m) m*v1 +m*v2= 2m*v' we know the velocity after the collision equals 16.53 m/s m*v1 +m*v2= 2m*(16.53) Distribute the 2 to the velocity m*v1 +m*v2= m*(33.06) Factor out mass v1 + v2 = 33.06 That should get you nearly to the end. You have to find one of the cars velocities at the collision. I assume the major road is 45 mph. So using the same conversion factors in a.) the major car's velocity is 20.12 m/s. You can just plug that into the momentum equation because i believe the problem says they .hit at either an angle of 20 degrees or 70 degrees in which case the velocities of the cars will have to be adjusted for their trigonometric values. So depending on the angle of collision and or how you drew your diagram you would have the multiply the velocity of the major road by sin or cos of the angle depending on the triangle. For that you need the picture, so hopefully that gets you to the end. Good luck!