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Deceleration (Posted on 2003-10-02) Difficulty: 3 of 5
George is driving 100 ft/sec toward an intersection.
He looks to his right, and sees Bill, driving 30 ft/sec toward the same intersection. George foolishly slams on his brakes.

If he had kept going 100 ft/sec, he would have been through the intersection long before Bill got there.

At the instant that he slams on his brakes, the center of George's car is 125 ft from the intersection, and the center of Bill's car is 150 ft from the intersection. George's brakes give his car an acceleration of -30 ft/sec².
Bill never changes his speed.
Each car is 13 ft long and 7 ft wide.

Will there be a collision?

See The Solution Submitted by DJ    
Rating: 4.2308 (13 votes)

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Solution solution | Comment 1 of 16
First let's figure when the centers of the cars will pass the point where their (the centers) paths cross. It's assumed that the incident takes place in America, where people drive on the right side of the road, and that Bill, in his car, does not react in any way to the impending potential problem, while George's only reaction is in applying the brakes (neither swerves). We'll further assume that the left side of the car is on the center line of each respective road, as we are not told how much distance to allow or how wide the road is or in what lane; at least it's better than assuming driving down the center of the road altogether. The width of the highway is not given so it's assumed that the distance to the intersection is the distance to the intersecting roadway center line.

For this center-point calculation, Bill's car must travel 150-3.5 feet (to allow for the half car-width that George's car is on Bill's side of the center line of George's road) at 30 ft/s, or 4.883 seconds.

George's center has 125+3.5 feet to get to the car-center intersection point. To find the time to get there, recognize that the average speed for any duration of time from the application of the brakes is 100-15t ft/s, so the time is found via
(100-15t)t = 128.5
15t^2-100t+128.5=0
t=10/3 +/- 1.595 sec

The higher number of seconds would be if the negative acceleration continued after the car came to a stop, and started going backward into the intersection again, so the lesser time is what's sought, or 1.738 seconds. So George's car does pass the intersection before Bill's; but is is soon enough to avoid a collision?

George's car must clear (travel) a total of 125 + 7 + 13/2 feet to account for the full width of Bill's car and the back half of itself. That's 138.5 feet.

(100-15t)t = 138.5
15t^2-100t+138.5=0
t=10/3 +/- 1.370 sec

so the time is 1.963 seconds, by which time Bill's car has traveled 1.963 x 30 = 58.89 feet, bringing its front 150-58.89-7 feet back from the path of the right side of George's car, which is still a safe margin even if George is in a right lane or somewhat further to the right of the centerline than we allowed for.
  Posted by Charlie on 2003-10-02 09:23:29
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