MHB UP 2.1 Rocket Average Velocity

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    Rocket Velocity
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The discussion focuses on calculating the average velocity of a rocket during its ascent. For the 4.75 seconds after clearing the launch platform, the average velocity is determined to be approximately 197 m/s, calculated by subtracting the height of the launch platform from the rocket's altitude and dividing by the time. For the first 5.90 seconds of flight, the average velocity is approximately 169 m/s, calculated by dividing the total distance of 1.00 km by the total time of 5.90 seconds. The calculations utilize the definition of average velocity as distance traveled over time taken. These results highlight the rocket's acceleration during its initial flight phases.
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$\tiny{UP 2.1}$
$\textsf{A rocket carrying a satellite is accelerating straight up from the Earth's surface}$
$\textsf{At 1.15 after liftoff, the rocket clears the top of its launch platform, 63m above the ground.}$
$\textsf{After an additional 4.75s it is 1.00km above the ground.}$
$\textsf{Calculate the magnitude of the average velocity of the rocket for}$
$\textsf{a. the 4.75s part of its flight;}$
$\textsf{b. the first 5.90s of its flight.}$
$\textsf{answer: a. $197m/s$ b.$169m/s$}$

didn't know where to post this since didn't see a Physics category

asume this is done finding slope
 
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karush said:
$\tiny{UP 2.1}$
$\textsf{A rocket carrying a satellite is accelerating straight up from the Earth's surface}$
$\textsf{At 1.15 after liftoff, the rocket clears the top of its launch platform, 63m above the ground.}$
$\textsf{After an additional 4.75s it is 1.00km above the ground.}$
$\textsf{Calculate the magnitude of the average velocity of the rocket for}$
$\textsf{a. the 4.75s part of its flight;}$
$\textsf{b. the first 5.90s of its flight.}$
$\textsf{answer: a. $197m/s$ b.$169m/s$}$

didn't know where to post this since didn't see a Physics category

asume this is done finding slope
All you need here is the definition of average velocity: $$\text{average velocity} = \frac{\text{distance travelled}} {\text{time taken}}.$$ For a., the (vertical) distance is 1000m $-$ 63m, and the time taken is 4.75s. For b., the distance is 1000m and the time is 1.15s $+$ 4.75s.
 
Opalg said:
All you need here is the definition of average velocity: $$\text{average velocity} = \frac{\text{distance travelled}} {\text{time taken}}.$$ For a., the (vertical) distance is 1000m $-$ 63m, and the time taken is 4.75s. For b., the distance is 1000m and the time is 1.15s $+$ 4.75s.

a.
the 4.75s part of its flight;
\begin{align*}\displaystyle
&=\frac{1000-63}{4.75}\approx \color{red}{197 \, m/s}
\end{align*}
b. the first 5.90s of its flight
\begin{align*}\displaystyle
&=\frac{1000}{5.9}\approx \color{red}{169 \, m/s}
\end{align*}
 
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