Math / Algebra

QuestionIf an arrow is shot straight upward on the moon with a velocity of $50 \mathrm{~m} / \mathrm{s}$ , its height (in meters) after t seconds is given by $s(t)=50 t-0.83 t^{2}$ . What is the velocity of the arrow (include units) after 7 seconds? $\square$ How long will it take for the arrow to return and hit the moon (include units)? $\square$ With what velocity (include units) will the arrow hit the moon? $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? If an arrow goes up and comes down, how fast is it going at 7 seconds, when does it land, and how fast is it going when it lands? Watch out! Velocity can be positive or negative, make sure to keep track of the direction!

STEP 2

1. Find the velocity function.
2. Calculate the velocity after 7 seconds.
3. Find when the arrow lands.
4. Calculate the landing velocity.

STEP 3

Alright, so we're given the height function, $s(t) = 50t - 0.83t^2$ , which tells us how high the arrow is at any time $t$ .
But we need the velocity function!

STEP 4

Remember, velocity is the rate of change of position.
That's a fancy way of saying it's the derivative of the height function.
So, let's find the derivative!

STEP 5

$v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} (50t - 0.83t^2)$

STEP 6

Using the power rule, the derivative of $50t$ is just 50, and the derivative of $-0.83t^2$ is $-0.83 \cdot 2t = -1.66t$ .

STEP 7

So, our velocity function is $v(t) = 50 - 1.66t$ .
Awesome!

STEP 8

Now, we just need to plug in $t = 7$ into our velocity function:

STEP 9

$v(7) = 50 - 1.66 \cdot 7 = 50 - 11.62 = 38.38$

STEP 10

So, after 7 seconds, the arrow is traveling at $\bold{38.38 \ m/s}$ .
Still going up!

STEP 11

The arrow lands when its height is zero, right?
So, we need to solve $s(t) = 0$ :

STEP 12

$50t - 0.83t^2 = 0$

STEP 13

We can factor out a $t$ : $t(50 - 0.83t) = 0$

STEP 14

This gives us two solutions: $t = 0$ and $50 - 0.83t = 0$ . $t = 0$ is the initial time, when the arrow is launched.
We're interested in when it lands, so let's solve for the other $t$ :

STEP 15

$50 = 0.83t$ $t = \frac{50}{0.83} \approx 60.24$

STEP 16

So, the arrow lands after approximately $\bold{60.24 \ seconds}$ .

STEP 17

Now, we just need to find the velocity at $t = 60.24$ :

STEP 18

$v(60.24) = 50 - 1.66 \cdot 60.24 = 50 - 99.9984 \approx -50$

STEP 19

The landing velocity is approximately $\bold{-50 \ m/s}$ .
The negative sign means it's going downwards, which makes sense!

STEP 20

The velocity after 7 seconds is $38.38 \ m/s$ .
The arrow lands after $60.24 \ seconds$ .
The landing velocity is $-50 \ m/s$ .

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Studdy Solution

STEP 1

What is this asking? If an arrow goes up and comes down, how fast is it going at 7 seconds, when does it land, and how fast is it going when it lands? Watch out! Velocity can be positive or negative, make sure to keep track of the direction!

STEP 2

1. Find the velocity function.2. Calculate the velocity after 7 seconds.3. Find when the arrow lands.4. Calculate the landing velocity.

STEP 3

Alright, so we're given the height function, s(t)=50t−0.83t2s(t) = 50t - 0.83t^2s(t)=50t−0.83t2, which tells us how high the arrow is at any time ttt.But we need the *velocity* function!

STEP 4

Remember, velocity is the **rate of change** of position.That's a fancy way of saying it's the derivative of the height function.So, let's **find the derivative**!

STEP 5

v(t)=ddts(t)=ddt(50t−0.83t2)v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} (50t - 0.83t^2)v(t)=dtd​s(t)=dtd​(50t−0.83t2)

STEP 6

Using the power rule, the derivative of 50t50t50t is just **50**, and the derivative of −0.83t2-0.83t^2−0.83t2 is −0.83⋅2t=−1.66t-0.83 \cdot 2t = -1.66t−0.83⋅2t=−1.66t.

STEP 7

So, our **velocity function** is v(t)=50−1.66tv(t) = 50 - 1.66tv(t)=50−1.66t.Awesome!

STEP 8

Now, we just need to **plug in** t=7t = 7t=7 into our velocity function:

STEP 9

v(7)=50−1.66⋅7=50−11.62=38.38v(7) = 50 - 1.66 \cdot 7 = 50 - 11.62 = 38.38v(7)=50−1.66⋅7=50−11.62=38.38

STEP 10

So, after 7 seconds, the arrow is traveling at 38.38 m/s\bold{38.38 \ m/s}38.38 m/s.Still going up!

STEP 11

The arrow lands when its height is zero, right?So, we need to **solve** s(t)=0s(t) = 0s(t)=0:

STEP 12

50t−0.83t2=050t - 0.83t^2 = 050t−0.83t2=0

STEP 13

We can factor out a ttt: t(50−0.83t)=0t(50 - 0.83t) = 0t(50−0.83t)=0

STEP 14

This gives us two solutions: t=0t = 0t=0 and 50−0.83t=050 - 0.83t = 050−0.83t=0. t=0t = 0t=0 is the **initial time**, when the arrow is launched.We're interested in when it lands, so let's solve for the other ttt:

STEP 15

50=0.83t50 = 0.83t50=0.83t t=500.83≈60.24t = \frac{50}{0.83} \approx 60.24t=0.8350​≈60.24

STEP 16

So, the arrow lands after approximately 60.24 seconds\bold{60.24 \ seconds}60.24 seconds.

STEP 17

Now, we just need to find the velocity at t=60.24t = 60.24t=60.24:

STEP 18

v(60.24)=50−1.66⋅60.24=50−99.9984≈−50v(60.24) = 50 - 1.66 \cdot 60.24 = 50 - 99.9984 \approx -50v(60.24)=50−1.66⋅60.24=50−99.9984≈−50

STEP 19

The landing velocity is approximately −50 m/s\bold{-50 \ m/s}−50 m/s.The negative sign means it's going *downwards*, which makes sense!

STEP 20

The velocity after 7 seconds is 38.38 m/s38.38 \ m/s38.38 m/s.The arrow lands after 60.24 seconds60.24 \ seconds60.24 seconds.The landing velocity is −50 m/s-50 \ m/s−50 m/s.

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1. Find the velocity function.
2. Calculate the velocity after 7 seconds.
3. Find when the arrow lands.
4. Calculate the landing velocity.

Alright, so we're given the height function, $s(t) = 50t - 0.83t^2$ , which tells us how high the arrow is at any time $t$ .
But we need the velocity function!

Remember, velocity is the rate of change of position.
That's a fancy way of saying it's the derivative of the height function.
So, let's find the derivative!

$v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} (50t - 0.83t^2)$

Using the power rule, the derivative of $50t$ is just 50, and the derivative of $-0.83t^2$ is $-0.83 \cdot 2t = -1.66t$ .

So, our velocity function is $v(t) = 50 - 1.66t$ .
Awesome!

Now, we just need to plug in $t = 7$ into our velocity function:

$v(7) = 50 - 1.66 \cdot 7 = 50 - 11.62 = 38.38$

So, after 7 seconds, the arrow is traveling at $\bold{38.38 \ m/s}$ .
Still going up!

The arrow lands when its height is zero, right?
So, we need to solve $s(t) = 0$ :

$50t - 0.83t^2 = 0$

We can factor out a $t$ : $t(50 - 0.83t) = 0$

This gives us two solutions: $t = 0$ and $50 - 0.83t = 0$ . $t = 0$ is the initial time, when the arrow is launched.
We're interested in when it lands, so let's solve for the other $t$ :

$50 = 0.83t$ $t = \frac{50}{0.83} \approx 60.24$

So, the arrow lands after approximately $\bold{60.24 \ seconds}$ .

Now, we just need to find the velocity at $t = 60.24$ :

$v(60.24) = 50 - 1.66 \cdot 60.24 = 50 - 99.9984 \approx -50$

The landing velocity is approximately $\bold{-50 \ m/s}$ .
The negative sign means it's going downwards, which makes sense!

The velocity after 7 seconds is $38.38 \ m/s$ .
The arrow lands after $60.24 \ seconds$ .
The landing velocity is $-50 \ m/s$ .