# A rocket with a satellite accelerates upward from the earth's surface. At 1.15 sec. after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. What is the magnitude of the average velocity of the rocket for the 4.75 s flight?

##### 2 Answers
Jul 26, 2015

The average velocity of the rocket during the first 4.75s is 226.75m/s upwards.

#### Explanation:

Note: I'm assuming a constant acceleration for this problem.

We can solve this problem using basic kinematics.

Since the rocket was at rest when launched, our initial velocity is $0 \frac{m}{s}$ and we'll define the upwards direction as positive.

Find the acceleration of the rocket using our first set of info
$d = {v}_{i} t + \frac{1}{2} a {t}^{2}$
$a = 2 \frac{d - {v}_{i} t}{t} ^ 2 = 2 \frac{63 m - 0}{1.15 s} ^ 2 \approx 95.27 \frac{m}{s} ^ 2$

Find the final velocity of the rocket at 4.75s
${v}_{f} = {v}_{i} + a t = 0 + \left(95.27 \frac{m}{s} ^ 2\right) \left(4.75 s\right) \approx 452.55 \frac{m}{s}$

Find the average velocity of the rocket for the first 4.75s
$\overline{v} = \frac{{v}_{i} + {v}_{f}}{2} = \frac{0 + 452.55 \frac{m}{s}}{2} \approx 226.75 \frac{m}{s}$

Since we're looking for an average velocity and not an average speed we also need to include a direction, so our final answer is $226.75 \frac{m}{s}$ upwards.

Jul 27, 2015

$\overline{v} = \text{197 m/s}$

#### Explanation:

FULL QUESTION

A rocket with a satellite accelerates upward from the Earth's surface. At 1.15 sec. after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 sec, it is 1.00 km above the ground.

What is the magnitude of the average velocity of the rocket for the 4.75 s flight?

So, you know that your rocket covers 63 m in 1.15 s, which means that you can determine the speed with which it clears the platform by using

$h = {v}_{0} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$

Since the rocket started at rest, its initial velocity will be equal to zero. The acceleration of the rocket for the first 1.15 s of flight will be

$h = \frac{1}{2} \cdot a \cdot {t}^{2}$

a = (2h)/t^2 = (2 * "63 m")/(1.15""^2 "s"^2) = "95.3 m/s"""^2

The rocket's velocity after 1.15 s of flight is

${v}_{1} = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t + a \cdot t$

${v}_{1} = 95.3 \text{m"/"s"^color(red)(cancel(color(black)("2"))) * 1.15color(red)(cancel(color(black)("s"))) = "109.6 m/s}$

Now, you know that after an additional 4.75 seconds, the rocket is at a height of 1.00 km above the ground.

This means that it covered, in 4.75 seconds of flight, an additional

${h}_{2} = 1000 - 63 = \text{937 m}$

Use this distance and the flight time to determine its acceleration for the second portion of the flight

${h}_{2} = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= {v}_{1}}} \cdot {t}_{2} + \frac{1}{2} \cdot {a}_{2} \cdot {t}_{2}^{2}$

${a}_{2} = \frac{2 \cdot \left({h}_{2} - {v}_{1} \cdot {t}_{2}\right)}{t} _ {2}^{2}$

a_2 = (2 * ("937 m" - 109.6"m"/color(red)(cancel(color(black)("s"))) * 4.75color(red)(cancel(color(black)("s")))))/(4.75^2"s"^2)

a_2 = (2 * "416.4 m")/("22.6 s"^2) = "36.8 m/s"""^2

This means tha the speed of the rocket after the second part of its flight will be

${v}_{2} = {v}_{1} + {a}_{2} \cdot {t}_{2}$

${v}_{2} = 109.6 \text{m"/"s" + 36.8"m"/color(red)(cancel(color(black)("s"))) * 4.75color(red)(cancel(color(black)("s"))) ="284.4 m/s}$

Therefore, the magnitude of the average velocity is

$\overline{v} = \frac{{v}_{1} + {v}_{2}}{2} = \frac{109.6 + 284.4}{2} = \textcolor{g r e e n}{\text{197 m/s}}$