A plane leaves with an acceleration of 6.34 m/s squared and takes 1.5 hours to stop. What is the speed of the plane? What was th

Question

2 years ago

5

e distance it traveled?

1 answer:

Maurinko [17] · Answer 1 · 2023-02-07T21:39:55+03:00

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:-

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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