The displacement vector for a 15.0 second interval of a jet airplane\'s flight i

Question

The displacement vector for a 15.0 second interval of a jet airplane's flight is (3950, ?2430) m. (a) What is the magnitude of the average velocity? (b) At what angle, measured from the positive x axis, did the airplane fly during this time interval? Express the angle as a number between ?180 The displacement vector for a 15.0 second interval of a jet airplane's flight is (3950, ?2430) m. (a) What is the magnitude of the average velocity? (b) At what angle, measured from the positive x axis, did the airplane fly during this time interval? Express the angle as a number between ?180

Explanation / Answer

a) In order to be able to get the magnitude of the average velocity, you need to know the displacement vector, which is the resultant vector of the x and y components of the airplane's flight. This vector can be easily calculated with the use of the pythagorean theorem:

resultant vector(r) = square root of { [x-component]^2 + [y-component]^2 }
r = sqrt [(3950 m)^2 + (-2430 m)^2]
r = sqrt [15602500 m^2 + 5904900 m^2]
r = sqrt [21507400 m^2]
r = 4637.60 m

the displacement vector is approx. 4637.60 m. The magnitude of the average velocity can now be computed by dividing the resultant vector to the time interval given.

ave. velocity(v) = displacement vector / time interval
v = 4637.60 m / 15.0 s
v = 220.78 m/s

magnitude of average velocity = 309.17 m/s (considering the number of significant figures)

b) the angle can be calculated simply with the use of trigonometric functions with the aid of a scientific calculator.

tangent (angle) = opposite side / adjacent side
angle = arc tangent of [(-2430 m)/(3950 m)]
angle = arctan[-0.615]
angle = -31.6 degrees

the angle is -31.6 degrees from the positive x-axis

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.