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11. [-/4 Points] DETAILS MY NOTES SCALCET9 2.7.012.MI. If a rock is thrown upward on the planet Mars with a velocity of \( 15 \mathrm{~m} / \mathrm{s} \), its height above the ground (in meters) a \( t \) seconds is given by \( \mathrm{H}=15 \mathrm{t}-1.86 \mathrm{t}^{2} \). (a) Find the velocity (in \( \mathrm{m} / \mathrm{s} \) ) of the rock after 2 seconds. (b) Find the velocity (in \( \mathrm{m} / \mathrm{s} \) ) of the rock when \( t=a \). (c) When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.) ( t . \( \mathrm{m} / \mathrm{s} \) (d) With what velocity (in \( \mathrm{m} / \mathrm{s} \) ) will the rock hit the surface? (d \( \mathrm{m} / \mathrm{s} \)

Ask by Murphy Love. in the United States
Mar 10,2025

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**(a) After 2 seconds, the velocity is 7.56 m/s.** **(b) When \( t = a \), the velocity is \( 15 - 3.72a \) m/s.** **(c) The rock hits the surface at approximately 8.1 seconds.** **(d) When the rock hits the surface, its velocity is -15 m/s.

Solución

**(a) Finding the velocity after 2 seconds** The height function is given by \[ H(t)=15t-1.86t^2. \] The velocity is the derivative of the height with respect to time: \[ v(t)=H'(t)=\frac{d}{dt}(15t-1.86t^2)=15-3.72t. \] At \( t=2 \) seconds, \[ v(2)=15-3.72(2)=15-7.44=7.56 \, \text{m/s}. \] **(b) Finding the velocity when \( t=a \)** Using the velocity function, \[ v(t)=15-3.72t, \] when \( t=a \) the velocity is \[ v(a)=15-3.72a. \] **(c) Determining when the rock hits the surface** The rock hits the surface when \( H(t)=0 \): \[ 15t-1.86t^2=0. \] Factor out \( t \): \[ t(15-1.86t)=0. \] This gives two solutions: \[ t=0 \quad \text{or} \quad 15-1.86t=0. \] We discard \( t=0 \) because it represents the initial time. Solving for \( t \) in the second equation: \[ 1.86t=15 \quad \Rightarrow \quad t=\frac{15}{1.86}. \] Calculating the value, \[ t\approx8.0645 \, \text{seconds}. \] Rounded to one decimal place, the rock hits the surface at approximately \[ t\approx8.1 \, \text{seconds}. \] **(d) Finding the velocity when the rock hits the surface** We use the velocity function: \[ v(t)=15-3.72t. \] Substitute \( t=\frac{15}{1.86} \) into the equation: \[ v\left(\frac{15}{1.86}\right)=15-3.72\left(\frac{15}{1.86}\right). \] Notice that \[ \frac{3.72}{1.86}=2, \] so the equation simplifies to \[ v\left(\frac{15}{1.86}\right)=15-2(15)=15-30=-15 \, \text{m/s}. \]

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Beyond the Answer

To dive into the rock's journey on Mars, let's first calculate its velocity. The velocity of the rock at any time \( t \) can be found by taking the derivative of the height function \( H \) with respect to \( t \): **(a)** The height function is \( H = 15t - 1.86t^2 \). The velocity function \( V(t) \) is \( V(t) = \frac{dH}{dt} = 15 - 3.72t \). Plugging in \( t = 2 \) seconds, we have: \[ V(2) = 15 - 3.72(2) = 15 - 7.44 = 7.56 \, \text{m/s} \] **(b)** For the general case at \( t = a \), the velocity is: \[ V(a) = 15 - 3.72a \] **(c)** To determine when the rock hits the surface, set \( H = 0 \): \[ 0 = 15t - 1.86t^2 \] Factoring gives us: \[ t(15 - 1.86t) = 0 \] So, \( t = 0 \) (launch time) or \( t = \frac{15}{1.86} \approx 8.1 \) seconds. **(d)** Finally, at \( t = 8.1 \) seconds, the velocity is: \[ V(8.1) = 15 - 3.72(8.1) \approx 15 - 30.12 = -15.12 \, \text{m/s} \] (The negative indicates it's moving downward.) So, in summary: - After 2 seconds, the velocity is **7.56 m/s**. - The velocity at \( t = a \) is **\( 15 - 3.72a \, \text{m/s}**. - The rock hits the surface at **8.1 seconds**. - The rock hits the surface with a velocity of **-15.12 m/s**.

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