Maybe one of the easier ways to work this is to write the equation of the line through the points you know, then use that to find the point you don't know. The 2-point formula for a line is ...
... y = (y2 -y1)/(x2 -x1)·(x -x1) +y1
Filling in (x1, y1) = (3, 213) and (x2, y2) = (8, 568), you get ...
... y = (568 -213)/(8 -3)·(x -3) +213
... y = 71(x -3) +213 . . . . . . simplified a bit
Now, for x=5, the value of <em>a</em> is ...
... a = 71(5 -3) +213 = 142 +213 = 355
The value of <em>a</em> that results in a constant rate of change is 355.
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The equation can actually be simplified further to ...
... y = 71x
(a) Given the position function
x(t) = (B m/s²) t² + 5 m
it's clear that the object accelerates at B m/s² (differentiate x(t) twice with respect to t), so that the force exerted on the object is
F(t) = (2 kg) (B m/s²) = 2B N
(b) Recall the work-energy theorem: the total work performed on an object is equal to the change in the object's kinetic energy. The object is displaced by
∆x = x(5 s) - x(0 s)
∆x = ((B m/s²) (5 s)² + 5 m) - ((B m/s²) (0 s)² + 5 m)
∆x = 25B m
Then the work W performed by F (provided there are no other forces acting in the direction of the object's motion) is
W = (2B N) (25B m) = 50B² J = 200 J
Solve for B :
50B² = 200
B² = 4
B = ± √4 = ± 2
Since the change in kinetic energy and hence work performed by F is positive, the sign of B must also be positive, so B = 2 and the object accelerates at 2 m/s².
(c) We found in part (b) that the object is displaced 25B m, and with B = 2 that comes out to ∆x = 50 m.
There are 342 inches, 28.5 feet, and 9.5 yards
Step-by-step explanation:
The presence of a midpoint will result in congruent segment.
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