The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.
Three equation of motion are:-
- v = u + at
- s = ut + (1/2)at²
- v² - u² = 2as
Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.
In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.
Applying equation 1 to find the initial speed of plane
v = u + at
0 = u + (-6.34 × 5400) {v=0 as plane will stop after 5400 sec}
u = 6.34 × 5400
u = 34236 m/s
Initial velocity of plane is 34236 m/s
Applying equation 2 to find the displacement of plane in that time period
s = ut + (1/2)at²
s = ( 34236 × 5400 ) - ( (1/2) × 6.34 × 5400² )
s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )
s = 5400 × ( 34236 - 17118 )
s = 5400 × 17118 metres
s = 5.4 × 17118 Km
s = 92437.2 Km
Distance travelled by plane is 92437.2 Km
So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.
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