(4 pts) If a rock is thrown vertically upward from the surface of Mars with an initial velocity of 15m/sec
1 answer:
Answer:
Step-by-step explanation:
I see you're in college math, so we'll solve this with calculus, since it's the easiest way anyway.
The position equation is
That equation will give us the height of the rock at ANY TIME during its travels. I could find the height at 2 seconds by plugging in a 2 for t; I could find the height at 12 seconds by plugging in a 12 for t, etc.
The first derivative of position is velocity:
v(t) = -3.72t + 15 and you stated that the rock will be at its max height when the velocity is 0, so we plug in a 0 for v(t):
0 = -3.72t + 15 and solve for t:\
-15 = -3.72t so
t = 4.03 seconds. This is how long it takes to get to its max height. Knowing that, we can plug 4.03 seconds into the position equation to find the height at 4.03 seconds:
s(4.03) = -1.86(4.03)² + 15(4.03) so
s(4.03) = 30.2 meters.
Calculus is amazing. Much easier than most methods to solve problems like this.
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For D, it's like C, but you dont have the chart. Use subtraction to find the slope. Change in y over change in x (which means subtract the y values and divide them by the difference in x values). So. You'd do 0 - 3 over -6 - 3 which equals -3 over -9, which simplifies to 1/3. And that's your slope! So to find the y-int just do the same thing you did in the last problem, plug in an ordered pair and the slope into the slope-intercept equation and solve as a one step equation (steps in picture). And there you have it, the slope which is 1/3, and the y-int which is 2.
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