If a rock is thrown upward on the planet Mars with a velocity of 17 m/s, its height (in meters) after t seconds is given by H = 17t − 1.86t^2.
(a) Find the velocity of the rock after one second.
(b) Find the velocity of the rock when t = a.
(c) When will the rock hit the surface? (Round your answer to one decimal place.)
(d) With what velocity will the rock hit the surface?
To find the velocity of the rock after one second, we can take the derivative of the height equation with respect to time, t.
(a) To find the velocity after one second, we substitute t = 1 into the derivative of the height equation:
H = 17t - 1.86t^2
v = dH/dt = d(17t - 1.86t^2)/dt = 17 - 3.72t
Substituting t = 1, we have:
v = 17 - 3.72(1)
v = 17 - 3.72
v ≈ 13.28 m/s
Therefore, the velocity of the rock after one second is approximately 13.28 m/s.
(b) To find the velocity of the rock when t = a, we substitute t = a into the derivative of the height equation:
v = 17 - 3.72t
Therefore, the velocity of the rock when t = a is given by:
v = 17 - 3.72a
(c) To find when the rock will hit the surface, we need to find the time at which the height, H, becomes zero. Substituting H = 0 into the height equation:
0 = 17t - 1.86t^2
This is a quadratic equation that can be solved using factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = -1.86, b = 17, and c = 0. Plugging in these values, we get:
t = (-17 ± √(17^2 - 4(-1.86)(0)))/(2(-1.86))
t = (-17 ± √(289))/(2*(-1.86))
t = (-17 ± 17)/(2*(-1.86))
We can simplify this to:
t = (-17 ± 17)/(-3.72)
Calculating both solutions:
t₁ = (-17 + 17)/(-3.72) ≈ 0.0 s
t₂ = (-17 - 17)/(-3.72) ≈ 8.60 s
From the time perspective, it doesn't make sense for the rock to hit the surface at t = 0.0 s, so we ignore that solution.
Therefore, the rock will hit the surface approximately 8.6 seconds after being thrown.
(d) To find the velocity at which the rock hits the surface, we substitute t = 8.6 into the velocity equation:
v = 17 - 3.72t
Substituting t = 8.6, we have:
v = 17 - 3.72(8.6)
v ≈ -16.53 m/s
Therefore, the rock will hit the surface with a velocity of approximately -16.53 m/s (negative indicating downward direction).
To find the answers to the given questions, we need to understand the terms and equations provided and apply the appropriate calculations. Let's break down each question step by step.
(a) To find the velocity of the rock after one second, we need to find the derivative of the height equation with respect to time, t.
Taking the derivative of H = 17t − 1.86t^2 with respect to t, we get:
dH/dt = 17 - 2 * 1.86t
Now substitute t = 1 into the derivative equation:
dH/dt = 17 - 2 * 1.86 * 1 = 17 - 3.72 ≈ 13.28 m/s
Therefore, the velocity of the rock after one second is approximately 13.28 m/s.
(b) To find the velocity of the rock when t = a, substitute a into the derivative equation:
dH/dt = 17 - 2 * 1.86 * a = 17 - 3.72a
The velocity of the rock when t = a is given by the equation 17 - 3.72a.
(c) To find when the rock will hit the surface, we need to solve the equation H = 0, as the height of the rock would be zero when it hits the surface.
Substitute H = 0 into the height equation:
0 = 17t − 1.86t^2
Rearranging the equation:
1.86t^2 - 17t = 0
Now solve for t by factoring or applying the quadratic formula:
t(1.86t - 17) = 0
Either t = 0 or 1.86t - 17 = 0.
Solving 1.86t - 17 = 0:
1.86t = 17
t ≈ 9.14 seconds
Therefore, the rock will hit the surface approximately 9.14 seconds after it is thrown.
(d) To find the velocity at which the rock hits the surface, substitute t = 9.14 into the derivative equation:
dH/dt = 17 - 2 * 1.86 * 9.14 ≈ -34.18 m/s
Therefore, the rock will hit the surface with a velocity of approximately -34.18 m/s. The negative sign indicates that it is moving in the downward direction.
a) Well, after one second the velocity of the rock can be found by plugging in t = 1 into the equation H = 17t - 1.86t^2. So, let's do the math: H = 17(1) - 1.86(1)^2. After adding and multiplying, we get H = 17 - 1.86. That means the velocity of the rock after one second is 15.14 m/s. Don't worry, it's not rock-et science!
b) To find the velocity of the rock when t = a, we can simply substitute a into the equation H = 17t - 1.86t^2. Let me calculate this for you: H = 17(a) - 1.86(a)^2. After doing the calculations, we get H = 17a - 1.86a^2. So, the velocity of the rock when t = a is 17a - 1.86a^2 m/s. Pretty smashing, isn't it?
c) To find when the rock will hit the surface, we need to set the height (H) equal to zero and solve for t. Let's set up the equation: 0 = 17t - 1.86t^2. This equation is known as a quadratic equation, and we can solve it using the quadratic formula. After applying a little math, we find that the time it takes for the rock to hit the surface is approximately 9.1 seconds. That's quite a wait, but don't worry, time flies when you're having fun!
d) Lastly, to find the velocity at which the rock will hit the surface, we can substitute the time we found in part c (t = 9.1) into the equation H = 17t - 1.86t^2. Let's calculate it: H = 17(9.1) - 1.86(9.1)^2. After doing the calculations, we find that the velocity at which the rock will hit the surface is approximately -159.53 m/s. Woah, that's quite the drop! I guess we can say the rock will make a smashing entrance!
As on earth, v = Vo - at where a = 1.86*2 on Mars.
So, v = 17 - 3.72t
v(1) = 17-3.72 = 13.28m/s
v(a) = 17 - 3.72a
Since h = t(17-1.86t)
h=0 when t = 0 (rock thrown)
and t = 9.14 (rock hits again)
The rock will hit the surface with the same speed it had when it left: 17m/s. However, as a velocity, it's falling, so v = -17m/s