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An aircraft is flying at a height of 340...

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is `30^@`, what is the speed of the aircraft ?

Text Solution

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We are of the aircraft, h = 400 m
`angle AOB = 30 ^(@)`
We are to calculate : Speed of the aircraft.
Using Trigonometric relation, in right angledtriangle OAC,
`tan 15 ^(@) = ( AC)/(OC) `
`implies AC = OC tan 15 ^(@) = 3400 xx 0 .26 79 = 911 m`
As `15 ^(@) ` is covered in 5 s. So, speed of aircraft, ` v = ( 911)/(5) = 182. 2 ms ^(-1).`
Instantaneous acceleration : Instantaneous acceleration of an object is equal to limiting value of its average acceleration when time interval approaches zero
` vec a lim _( Delta t to 0) ( Delta vec v)/(Delta t) = ( vec (dv))/(dt)`
Direction of instantaneous acceleration : Instantaneous acceleration is along the direction in which change in velocity occurs.
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