An aircraft travelling at 600 km/h accelerates steadily at 10km/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',?
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The aircraft has to sound barrier ,i.e.
v=1100km/h=1100×5/18=2750/9 m/s
u=600km/h=600×5/18=1500/9 m/s
The acceleration of the aircraft 10km/h per second
a=10km/h/s=10×5/18 m/s/s= 25/9m/s^2
Using first equation of motion
v=u + at
2750/9=1500/9+ 25t/9
2750=1500+25t
27500-1500/25=1250/25
= 50s
hence the aircraft will cross the sound barrier in 50s.
v=1100km/h=1100×5/18=2750/9 m/s
u=600km/h=600×5/18=1500/9 m/s
The acceleration of the aircraft 10km/h per second
a=10km/h/s=10×5/18 m/s/s= 25/9m/s^2
Using first equation of motion
v=u + at
2750/9=1500/9+ 25t/9
2750=1500+25t
27500-1500/25=1250/25
= 50s
hence the aircraft will cross the sound barrier in 50s.
Alisha06:
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