Question

An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft's altitude , how long will it take to reach the 'sound barrier' ? ​

Answer 1

Given :

• Speed of aircraft = 600 km/h
• Speed of sound = 1100 km/ h
• Acceleration = 10 km/h/s

To find :

• Time to reach sound barrier

Conversions :

Speed of aircraft (u) = 600 km/ h

$= 600 \times \dfrac{5}{18}$

$= \sf \: \dfrac{1500}{9} \: ms {}^{ - 1}$

Speed of sound (v) = 1100 km/h

$= 1100 \times \dfrac{5}{18}$

$\sf = \dfrac{2750}{9} m{s}^{ - 1}$

acceleration of aircraft (a) = 10 km/h/s

$= 10 \times \dfrac{5}{18}$

$= \sf \dfrac{25}{9} m{s}^{ - 2}$

Using the first equation of motion

$\boxed{ \huge \sf \orange{v = u + at}}$

$\sf \implies \dfrac{2750}{9} = \dfrac{1500}{9} + \dfrac{25}{9} t$

$\sf \implies \: 2750 = 1500 + 25t$

$\sf \implies \: 2750 - 1500 = 25t$

$\sf \implies \: 1250 = 25t$

$\sf \implies \: t = \dfrac{1250}{25}$

$\boxed{ \large \bf{ \red {t = 50 \: s}}}$

$\underline{\rm \therefore The\ aircraft\ takes\ 50} \\ \rm\underline{\ seconds\ to\ reach\ the\ sound\ barrier}$

Answer 2

Answer:

$\huge\underline\bold\red{AnsWeR}$

• Answer. We have v = u + at. Therefore t = (v – u)/a = [(1100 – 600) km/h ]/ (10 km/h/s) = 50 s.