An accident occurred in which Driver 1 turned right from Post Road onto the eastbound lane of Route 460. Driver 2 was ap
Posted: Mon Jul 18, 2022 6:04 am
An accident occurred in which Driver 1 turned right from PostRoad onto the eastbound lane of Route 460. Driver 2 was approachingthis same intersection on Route 460, traveling east. Driver 2swerved left to avoid hitting Driver 1, passing driver 1, and thenstopping ahead of driver 1. Driver 1 passed the now stopped car ofDriver 2 and pulled onto the shoulder off Route 360 east of Driver2. Police arrived, and details for the accident were gathered.
Site Evidence
Driver 1’s car stopped a distance of 325 feet east of Post Road.Driver 1 stopped 50 feet east of Driver 2’s car. Driver 1’s car wasa 2018 Jeep Grand Cherokee. Driver 2’s car was a 2018 HondaFit.
Road conditions were dry. The road at this point is straight andlevel. No other cars were affected by the incident.
Driver Statements
Driver 1’s version: Driver 1 said he pulled outwhile Driver 2 was still some distance west of the intersection andthat Driver 1 had accelerated to 50 MPH before Driver 2 swervedaround him, evidently going well over the speed limit for Route460, which is 60 MPH. Had Driver 2 not been speeding, he would havebeen able to slow and not swerve.
Driver 2’s version: Driver 2 said he was drivingat 60 MPH, and Driver 1 pulled onto Route 460 when Driver 2 wasonly two car lengths from the intersection, forcing Driver 2 toswerve to avoid a collision. Driver 1 had barely started toaccelerate and could not have been going 50 MPH when Driver 2 hadto swerve around Driver 1. He started his swerve one car lengthbehind Driver 1 and started to stop when he was one car length infront of Driver 1.
Additional Information:
It may be useful to do calculations in units of feet/secondrather than miles/ hour. The conversion rate is 1 MPH = 1.467 ft/sso 50 mph = 73.3 ft/s and 60 mph = 88 ft/s.
1. Calculate the acceleration rate of the Jeep GrandCherokee in feet/second/second or ft/s2.
2. Calculate how much time Driver 1 would take to reach50 mph (73.3 ft/s) while accelerating at the rate determined inpart 1. Remember that the acceleration rate represents how much thespeed increases each second.
3. Next we need to figure out how far the car wouldtravel while accelerating at this rate (part 1) for this amount oftime (part 2). You have the data you need. Find the right equationand solve.
4. Now it’s time to evaluate the two driver's stories.If driver 2 passed driver 1 after driver 1 accelerated to 50 mph(73.3 ft/s), he would have to have started his deceleration fartherdown the road from the intersection than the distance calculated inpart 3. Add the estimated stopping distance for driver 2’s car tothe result of part 3 above. What is this distance?
5. Which driver’s account do you believe andwhy?
Site Evidence
Driver 1’s car stopped a distance of 325 feet east of Post Road.Driver 1 stopped 50 feet east of Driver 2’s car. Driver 1’s car wasa 2018 Jeep Grand Cherokee. Driver 2’s car was a 2018 HondaFit.
Road conditions were dry. The road at this point is straight andlevel. No other cars were affected by the incident.
Driver Statements
Driver 1’s version: Driver 1 said he pulled outwhile Driver 2 was still some distance west of the intersection andthat Driver 1 had accelerated to 50 MPH before Driver 2 swervedaround him, evidently going well over the speed limit for Route460, which is 60 MPH. Had Driver 2 not been speeding, he would havebeen able to slow and not swerve.
Driver 2’s version: Driver 2 said he was drivingat 60 MPH, and Driver 1 pulled onto Route 460 when Driver 2 wasonly two car lengths from the intersection, forcing Driver 2 toswerve to avoid a collision. Driver 1 had barely started toaccelerate and could not have been going 50 MPH when Driver 2 hadto swerve around Driver 1. He started his swerve one car lengthbehind Driver 1 and started to stop when he was one car length infront of Driver 1.
Additional Information:
It may be useful to do calculations in units of feet/secondrather than miles/ hour. The conversion rate is 1 MPH = 1.467 ft/sso 50 mph = 73.3 ft/s and 60 mph = 88 ft/s.
1. Calculate the acceleration rate of the Jeep GrandCherokee in feet/second/second or ft/s2.
2. Calculate how much time Driver 1 would take to reach50 mph (73.3 ft/s) while accelerating at the rate determined inpart 1. Remember that the acceleration rate represents how much thespeed increases each second.
3. Next we need to figure out how far the car wouldtravel while accelerating at this rate (part 1) for this amount oftime (part 2). You have the data you need. Find the right equationand solve.
4. Now it’s time to evaluate the two driver's stories.If driver 2 passed driver 1 after driver 1 accelerated to 50 mph(73.3 ft/s), he would have to have started his deceleration fartherdown the road from the intersection than the distance calculated inpart 3. Add the estimated stopping distance for driver 2’s car tothe result of part 3 above. What is this distance?
5. Which driver’s account do you believe andwhy?