A hawk is flying horizontally Southward at an altitude of 176.0 m and at a speed

Question

A hawk is flying horizontally Southward at an altitude of 176.0  m and at a speed of 2.20  m/s when he spots a slow-flying finch ahead of his current location and below his current path. At time zero, the hawk starts a dive by controlling the contact forces on him by the air so as to produce an acceleration of 5.20  m/s2 in a direction 76.0  degrees below horizontally Southward for a time of 0.950 seconds.

B) What is the hawk's displacement during the 0.950-s interval? Give your answer as an ordered pair, with magnitude first, followed by a comma, then followed by the direction. Give the direction in terms of an angle measured below the horizontally Southward direction.

C) Starting at the end of the 0.950-s interval, the hawk dives at constant velocity for 8.5 s. What is the hawk's displacement during the 8.5-s interval? Give your answer as an ordered pair, with magnitude first, followed by a comma, then followed by the direction. Give the direction in terms of an angle measured below the horizontally Southward direction.

D) At the end of the 8.5-s interval, the hawk misses the finch and has to "put on the brakes". He does so by controlling the contact forces on him by the air so as to produce an acceleration of 3.519  m/s2 in a direction 78.244  degrees above horizontally Northward for a time of 1.5 seconds. What is the hawk's final velocity at the end of the 1.5-s interval?Give your answer as an ordered pair, with magnitude first, followed by a comma, then followed by the direction. Give the direction in terms of an angle measured below the horizontally Southward direction. If the direction is above horizontally South instead of below, use a negative angle.

Explanation / Answer

(A) v0 = 2.20 m/s(0i - j ) = - 2.20 j m/s

a = 5.20 (-cos76j - sin76k) = -1.26j - 5.05k m/s^2

displacement = v0 t + a t^2 /2

= (-2.20j x 0.950) + (-1.26 j - 5.05k)(0.950^2)/2

= -2.09j - 0.57j - 2.28k

= -2.66j - 2.28k

magnitude = sqrt(2.66^2 + 2.28^2) = 3.50 m

direction = tan^-1(2.28 / 2.66) = 40.6 deg

Ans: 3.50 m , 40.6 deg

(C) v = v0 + a t = -3.40j - 4.80k

dispalcement = v t = -28.9j - 40.8k

magnitude = sqrt(28.9^2 + 40.8^2) = 50 m

angle = tan^-1(4.80/3.40) = 55 deg

Ans: 50 m , 55 deg

(D) now a = 3.519 ( cos78.244j + sin78.244k)

a = 0.717 j + 3.45 k m/s^2

vf = v + a t

= ( -3.40j - 4.80k ) + (0.717 j + 3.45k)(1.5)

= (-3.40 + 1.076)j + (-4.80 + 5.175)k

= -2.3j + 0.4 k

magnitude = sqrt(2.3^2 + 0.4^2) = 2.33 m/s

direction = - tan^-1(0.4 / 2.3 ) =- 10 deg

Ans: 2.33 m/s , - 10 deg

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