If a rock is thrown upward on the planet Mars with a velocity of 
13 m/s,
 its height above the ground (in meters) after t seconds is given by 
H = 13t − 1.86t2.
(a)
Find the velocity (in m/s) of the rock after 1 second.
9.98
 m/s
(b)
Find the velocity (in m/s) of the rock when 
t = a.
  m/s
(c)
When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.)
t = 
 s
(d)
With what velocity (in m/s) will the rock hit the surface?
 m/s

Let's go through each part of the problem, one step at a time.

The height function of the rock is given as:

\[
H(t) = 13t - 1.86t^2
\]

To solve these problems, we will need to use derivatives since velocity is the derivative of the height function.

### Part (a): Find the velocity after 1 second.

Velocity \( v(t) \) is the first derivative of the height function \( H(t) \):

\[
v(t) = \frac{d}{dt} \left( 13t - 1.86t^2 \right) = 13 - 2(1.86)t = 13 - 3.72t
\]

Now, plug in \( t = 1 \) second to find the velocity:

\[
v(1) = 13 - 3.72(1) = 13 - 3.72 = 9.28 \, \text{m/s}
\]

The given answer is 9.98 m/s, which may be due to rounding inconsistencies or a slight error in the velocity formula. According to our calculation, the velocity after 1 second is **9.28 m/s**.

### Part (b): Find the velocity when \( t = a \).

Using the velocity equation:

\[
v(t) = 13 - 3.72t
\]

The velocity at \( t = a \) seconds is:

\[
v(a) = 13 - 3.72a
\]

This is the general formula for velocity when \( t = a \).

### Part (c): When will the rock hit the surface?

The rock will hit the surface when the height \( H(t) = 0 \). Set the height equation to 0 and solve for \( t \):

\[
0 = 13t - 1.86t^2
\]

Factor the equation:

\[
0 = t(13 - 1.86t)
\]

This gives two solutions:

1. \( t = 0 \) (the time when the rock is initially thrown)
2. \( 13 - 1.86t = 0 \), which gives \( t = \frac{13}{1.86} \)

Solving for \( t \):

\[
t = \frac{13}{1.86} \approx 6.99 \, \text{seconds}
\]

So, the rock will hit the surface after approximately **7.0 seconds**.

### Part (d): With what velocity will the rock hit the surface?

Use the velocity equation to find the velocity when \( t = 6.99 \) seconds:

\[
v(6.99) = 13 - 3.72(6.99)
\]

\[
v(6.99) = 13 - 25.98 \approx -12.98 \, \text{m/s}
\]

The negative sign indicates that the rock is moving downward. Therefore, the velocity when the rock hits the surface is approximately **12.98 m/s** downward.

### Summary of Answers:

- (a) The velocity after 1 second is **9.28 m/s** (downward).
- (b) The velocity at \( t = a \) is \( 13 - 3.72a \, \text{m/s} \).
- (c) The rock will hit the surface after **7.0 seconds**.
- (d) The velocity when the rock hits the surface is **12.98 m/s** (downward).

Would you like to see further details, or do you have any questions?

### 5 Related Questions:
1. What is the rock's maximum height and at what time does it occur?
2. How would the height function change if the initial velocity were different?
3. How does the velocity of the rock change with time? Can you graph it?
4. What would happen to the height equation if gravity on Mars were different?
5. How do we determine the direction of motion from the velocity sign?

### Tip:
To find when an object reaches its maximum height, set the velocity function equal to zero and solve for time!

If a rock is thrown upward on the planet Mars with a velocity of 13 m/s, its height above the ground (in meters) after t seconds is given by H = 13t − 1.86t^2. (a) Find the velocity (in m/s) of the rock after 1 second. (b) Find the velocity (in m/s) of the rock when t = a. (c) When (in seconds) will the rock hit the surface? (d) With what velocity (in m/s) will the rock hit the surface?

Find out how to solve a motion problem on Mars where a rock is thrown with an initial velocity of 13 m/s. Learn how to calculate the velocity after 1 second, the time it hits the ground, and its final impact speed using differentiation and quadratic equations. Suitable for Grades 10-12.

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