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Question: An aircraft travelling at \(700\,km/h\) accelerating steadily at \(10\,km/h\,{\text{per}}\,{\text{se...

An aircraft travelling at 700km/h700\,km/h accelerating steadily at 10km/hpersecond10\,km/h\,{\text{per}}\,{\text{second}}. Taking the speed of sound as 1100km/h1100\,km/h at the aircraft, how long will it take to reach the sound barrier?

Explanation

Solution

Hint- In order to find the time taken by the aircraft to cross the speed of sound we can use the first equation of motion which is given as
v=u+atv = u + at
Where u is the initial velocity, v is the final velocity, t is the time taken, a is the acceleration.
By substituting the given values in this equation, we can find the time taken.

Step by step solution:
It is given that the velocity of an aircraft is 700km/h700\,km/h initially. So, we have
u=700km/hu = 700\,km/h
Let us convert this value given in kilometre per hour to meter per second. We know that
1km=1000m1\,km = 1000\,m and 1h=3600s1\,h = 3600\,s
Thus 1km1h=1000m3600s=5m18s\dfrac{{1\,km}}{{1\,h}} = \dfrac{{1000\,m}}{{3600\,s}} = \dfrac{{5\,m}}{{18\,s}}
So to convert value given in km/hkm/h to m/sm/s we need to multiply by the value by 518\dfrac{5}{{18}}
Therefore ,
u=700×518m/s=17509m/su = 700\, \times \dfrac{5}{{18}}\,m/s = \dfrac{{1750}}{9}\,m/s
We need to find the time taken by the aircraft to cross the speed of sound . It is given that the speed of sound is 1100km/h1100\,km/h. This is the final velocity . so,
v=1100km/hv = 1100\,km/h
v=1100×518m/s=27509m/sv = 1100\, \times \dfrac{5}{{18}}\,m/s = \dfrac{{2750}}{9}\,m/s
Acceleration of the aircraft is 10km/hpersecond10\,km/h\,{\text{per}}\,{\text{second}} . that is
a=10km/hperseconda = 10\,km/h\,{\text{per}}\,{\text{second}}
a=10×518m/s/s = 259m/s2a = 10\, \times \dfrac{5}{{18}}\,{\text{m/s/s = }}\dfrac{{25}}{9}\,m/{s^2}
Now let us find the time taken to cross the speed of sound. For that we can use the first equation of motion which is given as
v=u+atv = u + at
Where u is the initial velocity , v is the final velocity , t is the time taken , a is the acceleration.
Let us substitute the given values in this equation and solve for t .
27509=17509+259t\dfrac{{2750}}{9} = \dfrac{{1750}}{9} + \dfrac{{25}}{9}\,t
2750=1750+25t\Rightarrow 2750 = 1750 + 25t
25t=1000\Rightarrow 25t = 1000
t=40s\therefore t = 40\,s
This is the time taken to cross the speed of sound.

Note: In this question the values of velocities are given in km/hkm/h and the acceleration is given in km/h/skm/h/s don’t substitute these values directly in the first equation of motion. First convert all values to the standard units and then find the value of time.