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| Deceleration (Posted on 2003-10-02) |
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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?
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Submitted by DJ
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Rating: 4.2308 (13 votes)
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Solution:
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The idea used to solve this is the familiar distance formula:
d = vit + ½at²
where vi is the initial velocity, a is the acceleration, and d is the distance traveled in t seconds.
If an object is moving at constant speed, the acceleration a is just 0, so the forumula simply becomes:
d = vt
Before I tell you the answer, let's look at the wrong answer:
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First, use the formula for constant speed to find out when Bill will reach the intersection. He is 150' away, moving at 30 fps, so it is simple enough to solve for the time:
d = vt
150 = 30 t
t = 5 sec
He will reach the intersection in 5 seconds.
Now, we need to find out where George will be in 5 seconds.
He starts out with a speed (vi) of 100 fps, and his acceleration is -30 ft/s².
So, after 5 seconds (t=5):
d = vit - ½at²
d = 100(5) + ½(-30)(25)
d = 125
We see that George will have gone 125 feet, in those 5 seconds. In other words, he will be at the intersection at the same time as Bill. The dimentsions of the cars do not matter, the centers of their cars will exactly coincide. Crunch.
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But, it turns out, that is the wrong answer. There will be no collision. George will zoom right on through the intersection and screech to a halt before Bill even reaches it.
The discrepancy arises because, although George's brakes apply a constant deceleration, once he comes to a stop, he stops. If we assume the constant deceleration for the entire time, that would be the situation in which his car slows down, stops, and starts speeding up in reverse. Obviously, that is not the situation we are to consider here.
So, to solve this problem correctly, we first need to determine how long it took George to come to a complete stop.
We can do this with the formula for change in speed (with constant accelaration, and determine what time (t) it took him to go from vi=100 ft/s to a full stop (vf=0):
vf = vi + at
0 = 100 - 30t
t = 3 1/3 sec = 10/3
Then, we can find out where he ends up, but subsituting that time into the original equation:
d = vt - ½at²
d = 100(10/3) + ½(-30)(10/3)²
d = 1000/3 - (30/2)(100/9)
d = 1000/6 = 166 2/3
Subtract from that distance the 125' he had to travel, and he ends up stopping 41'8" beyond the intersection, 1.7 seconds before Bill crosses it. Even after considering the width and length of the cars, they are nowhere near each other.
There will be no collision.
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