Before I tell
you the answer, let's look at the wrong answer: Yes, there will be a collision.
Bill will be at the intersection in 5 sec.Return to my Puzzle pages
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© Copyright 1996, Jim Loy
Before I tell
you the answer, let's look at the wrong answer: Yes, there will be a collision.
Bill will be at the intersection in 5 sec.
d = vt
150 = 30 t
t = 5 sec
Where will George be in 5 sec?
d = vt - 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 centers of their cars will coincide. Crunch.
But, it turns out, that is the wrong answer. There will be no collision. George will zoom right on through the intersection, and will come to a halt 41 2/3 feet beyond the intersection, 1 2/3 seconds before Bill gets there.
How long will it take George to come to a stop?
v(final) = v(initial) - at
0 = 100 - 30t
t = 3 1/3 sec
Where will he end up?
d = vt - ½at²
d = 166 2/3
That's 41 2/3 feet beyond the intersection.
There will be no collision.
Where did the wrong answer come from? Well, George's distance formula is a quadratic equation, with two solutions for most distances. But, his brakes stop accelerating his vehicle when he comes to a stop. So, the solution that puts him at the intersection after 5 seconds, never occurs, as it would have happened after he came to a stop.
I used ft./sec. because other units may be so familiar that the solution can be guessed.