Question

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

Answer 1

Hint: Initial velocity of the aircraft, velocity of sound and acceleration of the aircraft are given in the question. We need to find time at which aircraft cross sound barriers. We can use the first equation of motion here. Here, aircraft need to reach the velocity of sound to cross the barrier. Hence, taking the velocity of sound as the final velocity we can find the time taken by the aircraft to cross the barrier.
Formula used:
\[v=u+at\]

Complete answer:
Given that,
Velocity of aircraft\[=600km/h=600\times \dfrac{5}{18}=\dfrac{1500}{9}m/s\]
Velocity of sound \[=1100km/h=1100\times \dfrac{5}{18}=\dfrac{2750}{9}m/s\]
Acceleration of the aircraft \[=10km/h/s=10\times \dfrac{5}{18}=\dfrac{25}{9}m/{{s}^{2}}\]
To reach the sound barrier, aircraft should attain the velocity of sound. i.e., final velocity of aircraft should be\[1100km/h\].
Let the aircraft reach the sound barrier, in \[t\]seconds.
Then, we have,
\[v=u+at\] --------- 1
Where,
\[v\] is the final velocity
\[u\] is the initial velocity of the object
\[a\] is the acceleration of the object
\[t\] is the time
Here,
Final velocity,\[v=\dfrac{2750}{9}m/s\]
Initial velocity,\[u=\dfrac{1500}{9}m/s\]
Acceleration,\[a=\dfrac{25}{9}m/{{s}^{2}}\]
Substitute above values in equation 1, we get,
\[\dfrac{2750}{9}=\dfrac{1500}{9}+\dfrac{25}{9}\times t\]
\[2750=1500+25t\]
\[t=\dfrac{2750-1500}{25}=50s\]
Aircraft will take \[50s\] to reach the sound barrier.

Note:
The three equations of motion in kinematics describe the most fundamental concepts of motion of an object in motion. These three equations govern the motion of an object in 1D, 2D and 3D. These equations can easily be used to calculate expressions such as the position, acceleration or velocity, of an object at various times.