121616949-math.197

# 121616949-math.197 - total distance traveled you must ﬁnd...

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9.2 Distance, Velocity, Acceleration 183 Similarly, since the velocity is an anti-derivative of the acceleration function a ( t ), we have v ( t ) = v ( t 0 ) + Z t t 0 a ( u ) du. EXAMPLE 9.5 Suppose an object is acted upon by a constant force F . Find v ( t ) and s ( t ). By Newton’s law F = ma , so the acceleration is F/m , where m is the mass of the object. Then we first have v ( t ) = v ( t 0 ) + Z t t 0 F m du = v 0 + F m u fl fl fl fl t t 0 = v 0 + F m ( t - t 0 ) , using the usual convention v 0 = v ( t 0 ). Then s ( t ) = s ( t 0 ) + Z t t 0 v 0 + F m ( u - t 0 ) du = s 0 + ( v 0 u + F 2 m ( u - t 0 ) 2 ) fl fl fl fl t t 0 = s 0 + v 0 ( t - t 0 ) + F 2 m ( t - t 0 ) 2 . For instance, when F/m = - g is the constant of gravitational acceleration, then this is the falling body formula (if we neglect air resistance) familiar from elementary physics: s 0 + v 0 ( t - t 0 ) - g 2 ( t - t 0 ) 2 , or in the common case that t 0 = 0, s 0 + v 0 t - g 2 t 2 . Recall that the integral of the velocity function gives the net distance traveled. If you want to know the
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Unformatted text preview: total distance traveled, you must ﬁnd out where the velocity function crosses the t-axis, integrate separately over the time intervals when v ( t ) is positive and when v ( t ) is negative, and add up the absolute values of the diﬀerent integrals. For example, if an object is thrown straight upward at 19.6 m/sec, its velocity function is v ( t ) =-9 . 8 t + 19 . 6, using g = 9 . 8 m/sec for the force of gravity. This is a straight line which is positive for t < 2 and negative for t > 2. The net distance traveled in the ﬁrst 4 seconds is thus Z 4 (-9 . 8 t + 19 . 6) dt = 0 ,...
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