An aeroplane lands at 216km/h and stops after covering a runway of 2km. calculate the acceleration and time in which it come te rest .
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Given conditions ⇒
Initial velocity = 216 km/hr.
= 216 × 5/18 m/s.
= 60 m/s.
Final Velocity = 0
Distance covered by it = 2 km.
= 2000 m.
Using the Formula,
v² - u² = 2aS
(0)² - (60)² = 2a(2000)
-3600 = 4000a
∴ a = -36/40
⇒ a = -0.9 m/s².
Negative sign shows that the Aeroplane is producing the retardation. Hence, the Retardation produced by it is 0.9 m/s².
Now, Using the Formula,
v - u = at
0 - 60 + (-0.9)t
∴ -60 = 0.9t
⇒ t = 66.67 s.
Hence, the time taken by the Aeroplane to produce the Retardation of 0.9 m/s² is 66.67 seconds.
Hope it helps.
Initial velocity = 216 km/hr.
= 216 × 5/18 m/s.
= 60 m/s.
Final Velocity = 0
Distance covered by it = 2 km.
= 2000 m.
Using the Formula,
v² - u² = 2aS
(0)² - (60)² = 2a(2000)
-3600 = 4000a
∴ a = -36/40
⇒ a = -0.9 m/s².
Negative sign shows that the Aeroplane is producing the retardation. Hence, the Retardation produced by it is 0.9 m/s².
Now, Using the Formula,
v - u = at
0 - 60 + (-0.9)t
∴ -60 = 0.9t
⇒ t = 66.67 s.
Hence, the time taken by the Aeroplane to produce the Retardation of 0.9 m/s² is 66.67 seconds.
Hope it helps.
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