Answer:
at 2/3 seconds
Step-by-step explanation:
S1(t) = t³ + 2
Average speed, dS1/dt = 3t²
S2(t) = t²
Average speed, dS2/dt = 2t
The distance between the objects is
dS1/dt - dS2/dt
= 3t² - 2t
The time the distance between the two object is at minimum is when the distance is 0
That is, when
3t² - 2t = 0
t(3t - 2) = 0
t = 0 or 3t - 2 = 0
t = 0 or t = 2/3
Answer:
{xlx = -6, -1, 0, 3}
fyly=-7, 62, 1,9}
{x|x = -7, -6, -2, -1, 0, 1, 3, 9}
{yl y = -7, -6, -2, -1, 0, 1, 3, 9
Step-by-step explanation:
Answer:
2.33 Seconds
Step-by-step explanation:
The equation y = -6t^2 - 10t + 56 expresses the height of a ball at a given time, t, on the planet Mars. We are also given that the ball is thrown with a velocity of 10 feet per second with the initial height of 56 feet.
To find the amount of time to reach the ground, we can say that the time being found will be when the ball is on the ground, or when y = 0. So we simply set our equation to 0 and solve for t.
y = -6t^2 - 10t + 56
0 = -6t^2 - 10t + 56
0 = -1 (6t^2 + 10t + -56)
0 = -1 (3t - 7) (2t + 8)
(3t - 7) = 0 OR (2t + 8) = 0
3t = 7 OR 2t = -8
t = 7/3 OR t = -4
Since time will not be negative, we will want to choose the positive solution for this quadratic equation.
Hence, the amount of time for the ball to hit the ground will be 7/3 seconds or 2.33 seconds.
Cheers.
Answer:
c
Step-by-step explanation:
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