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the velocity v of a falling object is directly proportional to the time t of the fall. If, after 2 seconds, the velocity of the object is 64 ft per second, what will its velocity be after 3 seconds?

2006-12-03 18:22:37 · 5 answers · asked by Penny C 1 in Science & Mathematics Mathematics

5 answers

velocity is directly proportional to time t

lets consider a propotionality constant C

then V = C*t

after when t = 2, then V= 64, use this to find the constant C, which will come out to be 32

substitute this in the equation, you get V=32t

then after 3 seconds, V=32*3= 96 ft/second

2006-12-03 18:27:15 · answer #1 · answered by huma 2 · 0 0

Technically speaking, if the object is falling, its velocity should be negative....

The velocity of the object = (-32) t

at 1 sec, the velocity is -32 ft/sec
at 2 sec, the velocity is -64 ft/sec
at 3 sec, the velocity is -96 ft/sec

The second derivative suggests that the acceleration of the object is 32 ft/s/s (an object's acceleration due to gravity)

This supports the fact that for every additional second the object falls, its velocity increases -32 ft/sec

http://home.alltel.net/okrebs/page205.html

2006-12-04 03:09:20 · answer #2 · answered by J D 2 · 0 0

v is directly propotional to t
=>v=kt where k is a constant
given, v=64 ft/sec at t=2 sec

=>64=k(2)
=>k=32 ft
By second condition

v=32*(3)=96 ft/sec

2006-12-04 02:44:37 · answer #3 · answered by sushant 3 · 0 0

96 ft/sec

2006-12-04 02:48:10 · answer #4 · answered by Srinivas c 2 · 0 0

v=a*t
64=a*2
a=32
32*3=96

2006-12-04 02:50:14 · answer #5 · answered by Uber_Ninja 1 · 0 0

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