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 197 m/s, calculated using the formula for average velocity, which is the distance traveled divided by the time taken. For the first 5.90 seconds of flight, the average velocity is calculated to be 169 m/s. These calculations are based on the rocket's altitude of 1.00 km and the initial height of 63 m above the ground.

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karush
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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*}
 
Last edited:

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