# PhysicsUP 2.1 rocket velocity

#### karush

##### Well-known member
$\tiny{UP 2.1}$
$\textsf{A rocket carrying a satellite is accelerating straight up from the earths surface}$
$\textsf{At 1.15 after liftoff, the rocket clears the top of its launch platform, 63m above the ground.}$
$\textsf{After an additional 4.75s it is 1.00km above the ground.}$
$\textsf{Calculate the magnitude of the average velocity of the rocket for}$
$\textsf{a. the 4.75s part of its flight;}$
$\textsf{b. the first 5.90s of its flight.}$
$\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

#### Opalg

##### MHB Oldtimer
Staff member
$\tiny{UP 2.1}$
$\textsf{A rocket carrying a satellite is accelerating straight up from the earths surface}$
$\textsf{At 1.15 after liftoff, the rocket clears the top of its launch platform, 63m above the ground.}$
$\textsf{After an additional 4.75s it is 1.00km above the ground.}$
$\textsf{Calculate the magnitude of the average velocity of the rocket for}$
$\textsf{a. the 4.75s part of its flight;}$
$\textsf{b. the first 5.90s of its flight.}$
$\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.

#### karush

##### Well-known member
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: