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.