Aplane flies from base camp to lake a, a distance of 290 km at a direction of 20.0° north of east. after dropping off supplies, the plane flies to lake b, which is 170 km and 30.0° west of north from lake a. graphically determine the distance and direction from lake b to the base camp.

We can simply find components of the given directions and conveniently add them up to find the total distance.

290 km 20 deg north of east can be decomposed as :

290 x cos(20) EAST + 290 sin(20) NORTH

=272.5 E + 99.18 N

Now from this point ,

170 Km 30 deg west of north :

170 cos(30) N - 170 sin (30) E   [ east = - west]

= 147.22 N - 85 E

hence coordinates of lake b are

(272.5 - 85) E + (99.18 + 147.22)N

187.5 E +264.4 N

use right triangle property to find distance D =

D = 324.13 Km

and angle = atan(b/a) = 54.65 deg

hence lake b is at 324.13 Km 54.65 deg east of north

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s