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Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
In the future, please post the full problem with all included instructions. After doing a quick internet search, I found your problem listed somewhere else. It mentions two parts (a) and (b)
Part (a) asked for the equation of the line in y = mx+b form
That would be y = -2x+9
This is because each time y goes down by 2, x goes up by 1. We have slope = rise/run = -2/1 = -2. This indicates that the height of the candle decreases by 2 inches per hour. The slope represents the rate of change.
The initial height of the candle is the y intercept b value. So we have m = -2 and b = 9 lead us from y = mx+b to y = -2x+9
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Part (b) then asks you to graph the equation. Because this is a linear equation, it produces a straight line. We only need 2 points at minimum to graph any line. Let's plot (0,9) and (1,7) on the same xy grid. These two points are the first two rows of the table. Plot those two points and draw a straight line through them. The graph is below
