19_Cal_Solution of Calculus_6e

# 19_Cal_Solution of Calculus_6e - t = 1, as the automobile...

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0.5 1 1.5 2 2.5 20 40 60 80 100 120 t = 1, as the automobile cannot suddenly jump from one position to a completely diFerent position in an instant. Hence 45 = 75 + C , so that C = 30. Therefore x ( t )= ½ 45 t if 0 5 t 5 1; 75 t 30 if 1 <t 5 2 . To see the graph of x ( t ), plot in Mathematica x[t ]:=I f[t < 1,45 t,75 t 30 ] on the interval 0 5 t 5 2. C01S02.074: The graph of x ( t ) will consist of three straight-line segments (because of the constant speeds), the ±rst of slope 60 for 0 5 t 5 1, the second of slope zero for 1 5 t 5 1 . 5, and the third of slope 60 for 1 . 5 5 t 5 2 . 5. The ±rst pair must coincide when t = 1 and the second pair must coincide when t =1 . 5 because the graph of x ( t ) can have no discontinuities. So if we write x ( t ) = 60 for 0 5 t 5 1, we must have x ( t ) = 60 for 1 5 t 5 1 . 5. ²inally, x ( t )=60 t + C for some constant C if 1 . 5 5 t 5 2 . 5, but the latter must equal 60 when t =1 . 5, so that C = 30. Hence x (
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## This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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