Answer:49 sec
Step-by-step explanation:
Given
Maximum speed to reach is 183.58 mi/h
Length of course is 5 mi
acceleration rate is defined by 60mi/h in 4 sec
therefore acceleration(a)
To reach a speed of 183.58mi/h with an acceleration of
Using equation of motion
v=u+at
t=0.00339 hours
t=12.23 s reach maximum speed
To complete course it takes
t=0.01360 hour
or
9514 1404 393
Answer:
a) 12
b) y +5 = 12(x +2)
Step-by-step explanation:
The slope of a function is its derivative.
a) The slope of y = x^3 +3 at any point is y' = 3x^2. So, at the point where x=-2, the slope of the curve is ...
m = 3(-2)^2 = 12 . . . . slope at (-2, -5)
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b) The equation of the tangent line is most easily written using the point-slope form of the equation of a line.
y -k = m(x -h) . . . . . line with slope m through point (h, k)
y +5 = 12(x +2) . . . . line with slope 12 through point (-2, -5)
