Question

Two friends, Phyllis and Joey, arruve together to an airport before their respective flights are leaving.

1.50kn) (103) = 1500m 2.05 10% = 2050 together to an airport before their Phyllis a= 2.35 mlo? do 1500 m d2050 m Problem 3 (20 points): Two friends, Phyllis and Joey, arrive together to an airport bere respective flights are leaving. The two craft off at the same time from different of course!). Phyllis plane accelerates a 35 m atheading of 7's of won a 1.50 runway. Joey's plane accelerates at 1.95 m/sbt aheading of 2'N of Won 2.05 km of the aircraft take off exactly at the end of their runways, 2 = 2.33 MIA a) Find the velocities of takeoff for each aircraft. 5 26 2 1500 -2.35 m/s (t) (2.35) = 1500 Va= 1500m a = 1.95 mlet No: 59.38.n14 235 Ja t t= 1500 22 22A Joey 2.351 2050 0 .95 +- 12.15 -5638.30 VA = 2050 hl Find 2.42 25.2 b) Find the acceleration of joey with respect to Phyllis. vi 323 c) Find the velocity of Phyllis at takeoff relative to Joey. d) What additional information would you have to know to determine how far Phyllis and Joey were apart when the aircraft took off?

Answer 1

a)
Initial velocity of Phyllis, u = 0
Consider v as the final velocity.
Acceleration, a = 2.35 m/s²
Distance travelled, s = 1500 m
Using the formula, v² - u² = 2as
v² = 0 + 2 x 2.35 x 1500
= 7050
v = SQRT[7050]
= 83.96 m/s

Initial velocity of Joey, u' = 0
Consider v as the final velocity.
Acceleration, a' = 1.95 m/s²
Distance travelled, s' = 2050 m
Using the formula, v'² - u'² = 2a's'
v'² = 0 + 2 x 1.95 x 2050
= 7050
v' = SQRT[7995]
= 89.41 m/s

b)
ajp - acceleration of joey with respect to Phyllis
ajg - acceleration of joey with respect to ground = - 1.95 cos(25) $\hat{x}$ + 1.95 sin(25) $\hat{y}$
apg - acceleration of Phyllis with respect to ground = - 2.35 cos(47) $\hat{x}$ - 2.35 sin(47) $\hat{y}$
agp - acceleration of ground with respect to Phyllis = -apg

ajp = ajg + agp
= ajg - apg
= (- 1.95 cos(25) $\hat{x}$ + 1.95 sin(25) $\hat{y}$ ) - (- 2.35 cos(47) $\hat{x}$ - 2.35 sin(47) $\hat{y}$ ​​​​​​​)
= -0.16 $\hat{x}$ + 2.54 $\hat{y}$ ​
|ajp| = SQRT[(-0.16)² + (2.54)²]
= ​​​​​​2.55 m/s²

c)

Vpj - velocity of Phyllis with respect to Joey
Vpg - velocity of Phyllis with respect to ground = - 83.96 cos(47) $\hat{x}$ - 83.96 sin(47) $\hat{y}$ ​​​​​​​
Vjg - velocity of Joey with respect to ground = - 89.41 cos(25) $\hat{x}$ + 89.41 sin(25) $\hat{y}$
Vgj - velocity of ground with respect to Joey = -apg

Vpj = Vpg + Vgj
= Vpg - Vjg
= (- 83.96 cos(47) $\hat{x}$ - 83.96 sin(47) $\hat{y}$ ​​​​​​​) - (- 89.41 cos(25) $\hat{x}$ + 89.41 sin(25) $\hat{y}$ )
= 23.77 $\hat{x}$ + 99.20 $\hat{y}$ ​
|Vpj| = SQRT[(23.77)² + (99.20)²]
= 102 m/s

d)
We need the initial distance between the runways.