# What is momentary acceleration

## Content: Integration: distance, speed, acceleration

a) Determine the function \$ v \$, which gives the speed of the rocket after \$ t \$ seconds. (Note for the constant of integration: The initial speed \$ v (0) = v_0 = 0 \$)

Calculation tip:

Since we are looking for the function of the velocity \$ v (t) \$, we have to integrate the formula of the acceleration \$ a (t) \$ once. We substitute \$ 5.5 \$ (we take this from the specification) for \$ a (t) \$ and calculate \$ v (t) \$. It should also be noted that the constant of integration \$ c = v_0 \$ does not apply, since our initial speed (speed at time \$ t = 0 \$) is \$ 0 \$ and therefore \$ v (0) = 0 \$ must apply.

\$\$ v (t) = \ int a (t) \ cdot dt \$\$ \$\$ v (t) = \ int 5.5 \ cdot dt \$\$ \$\$ v (t) = 5.5 \ cdot t + v_0 \$\$ \$\$ v (t) = 5.5 \ cdot t \$\$

Answer: The speed function is \$ v (t) = 5.5 \ cdot t \$.

b) Use \$ v \$ to determine the path function \$ s \$, which indicates the path covered after \$ t \$ seconds.

Calculation tip:

Since we are looking for the function of the path \$ s \$, we have to integrate the formula of the speed \$ v (t) \$ once. We substitute \$ 5.5t \$ (we take this from the result of exercise a) for \$ v (t) \$ and thus calculate our result for \$ s (t) \$. It should also be noted that the constant of integration \$ c = s_0 \$ does not apply, since our distance covered is \$ 0 \$ at the beginning.

\$\$ s (t) = \ int v (t) \ cdot dt \$\$ \$\$ s (t) = \ int 5.5t \ cdot dt \$\$ \$\$ s (t) = 5.5 \ cdot \ frac {t ^ 2 } {2} + c \$\$ \$\$ s (t) = 2.75 \ cdot t ^ 2 \$\$

Answer: The function equation is \$ s (t) = 2.75 \ cdot t ^ 2 \$.

c) \$ 180 \$ seconds after the start, the solid fuel boosters are burned out and blown off (source: Wikipedia).

Calculate ...

- ... the height at which the rocket is at this point in time.

Calculation tip:

\$ s (t) \$ is the formula of the distance traveled. We therefore put in this formula for the time \$ t = 180 \$, since we want to know where the rocket is after \$ 180 \$ seconds.

\$\$ s (t) = 2.75 \ times t ^ 2 \$\$ \$\$ s (180) = 2.75 \ times 180 ^ 2 \$\$ \$\$ s (180) = 89 100 m \$\$

Answer: After \$ 180 \$ seconds, our rocket is at \$ 89 100m \$.  The \$ k \$ of the tangent indicates the speed.

- ... the speed of the rocket at this point.

Calculation tip:

\$ v (t) \$ is the formula that specifies the speed at time \$ t \$. We therefore put in this formula for the time \$ t = 180 \$, because we want to know how high the speed of the rocket is after \$ 180 \$ seconds. The result appears in \$ m / s \$. If you wanted the result in \$ km / h \$, you would have to calculate the result again \$ 3.6 \$.

\$\$ v (t) = 5.5 \ times t \$\$ \$\$ v (180) = 5.5 \ times 180 \$\$ \$\$ v (180) = 990 m / s \$\$

\$\$ 990 m / s \ rightarrow 3 564 km / h \$\$

Answer: After \$ 180 \$ seconds, our rocket reached a speed of \$ 3,564 km / h \$.

d) The rocket reaches geostationary orbit at an altitude of about \$ 1000 km \$ (source: Wikipedia). Determine the number of seconds it takes to reach that orbit.

Calculation tip:

The function \$ s \$ describes the distance covered. If we assume the unrealistic assumption that the rocket rises “bolt straight”, we have to calculate when \$ s (t) = 1,000 km = 1,000,000 m \$. It is important that \$ s (t) \$ specifies the path in \$ m \$!

\$\$ 1 000 000 = 2.75 \ cdot t ^ 2 \$\$ \$\$ t = 603.02 \$\$

Answer: After \$ 603.02 \$ seconds, our rocket reached \$ 1,000,000 meters.