# RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY in Visual Studio .NET Generation QR Code in Visual Studio .NET RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY

RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY
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Fig. 18-4 where s is measured in feet and t in seconds.3 Here s0 and v0 are, respectively, the position (height) and the velocity of the object at time t = 0. The instantaneous velocity v is obtained by differentiating (18.1), v= EXAMPLES
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(a) At t = 0, a rock is dropped from rest from the top of a building 256 feet high. When, and with what velocity, does it strike the ground With s0 = 256 and v0 = 0, (18.1) becomes s = 256 16t 2 and the time of striking the ground is given by the solution of 0 = 256 16t 2 16t 2 = 256 t 2 = 16 t = 4 seconds Since we are assuming that the motion takes place when t 0, the only solution is t = 4 seconds. The velocity equation (18.2) is v = 32t, and so, for t = 4, v = 32(4) = 128 feet per second the minus sign indicating that the rock is moving downward when it hits the ground.
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ds = v0 32t dt
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(18.2)
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the position is measured in meters, the equation reads s = s0 + v0 t 4.9t 2 .
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RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY
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algebra
x feet per second = 60x feet per minute = 60(60x) feet per hour 3600x miles per hour = 5280 15 = x miles per hour 22
(18.3)
3 For example, 128 feet per second = 15 (128) = 87 11 miles per hour. 22
(b) A rocket is shot vertically from the ground with an initial velocity of 96 feet per second. When does the rocket reach its maximum height, and what is its maximum height With s0 = 0 and v0 = 96, (18.1) and (18.2) become s = 96t 16t 2 At a maximum value, or turning point, v = 0. Hence, 0 = 96 32t 32t = 96 t=3 Thus, it takes 3 seconds for the rocket to reach its maximum height, which is s = 96(3) 16(3)2 = 288 16(9) = 288 144 = 144 feet (c) When does the rocket of part (b) hit the ground It suf ces to set s = 0 in the free-fall equation (18.1), 0 = 96t 16t 2 0 = 6t t 2 0 = t(6 t) from which t = 0 or t = 6. Hence, the rocket hits the ground again after 6 seconds. Notice that the rocket rose for 3 seconds to its maximum height, and then took 3 more seconds to fall back to the ground. In general, the upward ight from point P to point Q will take exactly the same time as the downward ight from Q to P. In addition, the rocket will return to a given height with the same speed (magnitude of the velocity) that it had upon leaving that height. [divide by 16] and v= ds = 96 32t dt
Solved Problems
18.1 A stone is thrown straight down from the top of an 80-foot tower. If the initial speed is 64 feet per second, how long does it take to hit the ground, and with what speed does it hit the ground
Here s0 = 80 and v0 = 64. (The speed is the magnitude of the velocity. The minus sign for v0 indicates that the object is moving downward.) Hence, s = 80 64t 16t 2 and v= ds = 64 32t dt
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RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY
The stone hits the ground when s = 0, 0 = 80 64t 16t 2 0 = t 2 + 4t 5 0 = (t + 5)(t 1) t + 5 = 0 or t = 5 or t 1=0 t=1 [divide by 16]
Since the time of fall must be positive, t = 1 second. The velocity v when the stone hits the ground is v(1) = 64 32(1) = 64 32 = 96 feet per second