Starting

The aircraft has a take-off speed of 310 km/h, a take-off weight of 16.5 t, and uses a 2 km take-off runway. Its cruising speed is 880 km/h and cruising altitude is 10 km.
a) Determine the aircraft's acceleration during take-off and the take-off time.
b) Determine the thrust of the engines during take-off.
c) Over what minimum distance does the aircraft reach cruising speed and altitude at that thrust F?

Final Answer:

a =  1.8538 m/s²
t =  46.4516 s
F =  30587.3843 N
s₃ =  16116.5453 m

Step-by-step explanation:

$$\begin{gathered} v_0=310\text{km/h}\to \text{m/s}=310:3.6\text{m/s}=86.11111\text{m/s} \\ m=16.5\text{t}\to \text{kg}=16.5\cdot 1000\text{kg}=16500\text{kg} \\ s=2\text{km}\to \text{m}=2\cdot 1000\text{m}=2000\text{m} \\ v=880\text{km/h}\to \text{m/s}=880:3.6\text{m/s}=244.44444\text{m/s} \\ h=10\text{km}\to \text{m}=10\cdot 1000\text{m}=10000\text{m} \\ s=\frac{1}{2}\cdot a\cdot t^2 \\ {v}_0=at \\ s=\frac{1}{2}\cdot v_0\cdot t \\ t=\frac{2\cdot s}{v_0}=\frac{2\cdot 2000}{86.1111}=\frac{1440}{31}\doteq 46.4516\text{s} \\ a=v_0/t=86.1111/46.4516\doteq 1.8538{\text{m/s}}^2 \\ \text{ Verifying Solution: } \\ {s}_2=\frac{1}{2}\cdot a\cdot t^2=\frac{1}{2}\cdot 1.8538\cdot 46.4516^2=2000\text{m} \end{gathered}$$

$F=m\cdot a=16500\cdot 1.8538=30587.3843\text{N}$

$$\begin{gathered} v=a\cdot t_3 \\ {t}_3=v/a=244.4444/1.8538\doteq 131.8626\text{s} \\ {s}_3=\frac{1}{2}\cdot a\cdot t_3^2=\frac{1}{2}\cdot 1.8538\cdot 131.8626^2=16116.5453\text{m} \end{gathered}$$

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