v2 = (0, 25.0)km/hr

→ D = v1×3hrs + v2 × 1.5hrs =Answer (105.9, 99) km

A ship which can move at the velocity of v0 = (12, 6) m/s (x-axis: the West to the East) on the lake. The ship is now on the river which flows to the West at the speed of 5 m/s. (12 points) a.Calculate the velocity of the ship on the river. vwater = (-5, 0)m/s, v = v0 + vwater =Answer (7, 6) m/s b.Calculate the speed of the ship on the river. |v| =Answer 9.22 m/s c.The ship started sailing at A which is 23 km west and 47 km south from the origin (0. 0). Find the location of the ship after sailing 5 hours. x = A + vt = (-23km, -47km) + (7, 6)m/s × 5 × 3600s = (-23km, -47km) + (126, 108)km = Answer (103, 61) km d.How far did the ship sail in the distance from the initial location. |D| = Answer 166.0km

A Boeing 747 airplane is flying at the elevation of 2,000 km with the constant speed of 250 m/s around 300 km away from Baton Rouge. a. How long does the airplane take to reach Baton Rouge with this speed? t = 300km / 250 m/s = 1200s = Answer 20 minutes.

b. Near Baton Rouge, the airplane needs to slow down to the landing speed of 150 m/s before it starts the landing gear. How long does the airplane take to get the landing speed, if it slows down with the constant acceleration of -2 m/s2 in the horizontal direction? 150m/s = 250m/s - 2m/s2 t → t = Answer 50s.

c. While the airplane is slowing down in the problem b) above, how long horizontal distance does this airplane fly? This calculation will give the distance from the airport where the airplane starts preparing landing. Distance = 250m/s×50s + ½ (-2m/s2) (50s)2 = 10000m = Answer 10.0 km. d. While the airplane is slowing down in the problem b) above, if the airplane lowers its elevation by free-falling, what would be the elevation before the landing gear? y = y0 + ½gt2 = 2000km + ½ (-9.8m/s2) (50s)2 = 2000 km - 12.25 km = Answer 1987.75 km. e. By using the results from b) and c), what is the actual distance including both horizontal and vertical distances that the airplane flies during this duration. In other words, what is the magnitude of the displacement vector for this duration. displacement = √((10.0 km)2 + (-12.25 km)2) = Answer 15.81 km. f. What is the angle of flight with respect to the horizontal plane for this duration. In other words, what is the direction of the displacement vector for this duration. θ = tan-1[(-12.25km)/(10.0 km)] = -50.77° = Answer 0.886 radians g.When the airplane lands on the ground, its velocity is 50 m/s. The length of the runway of the Baton Rouge airport (BTR) is 2 km or 2000 m. If the airplane slows down with a constant acceleration, what would be the minimum acceleration needed for this airplane to stop within the runway? Acceleration = [0 - (50m/s)2)] / (2×2000m) = Answer - 0.625m/s^2.

A displacement vector r has a magnitude of r = 175 m and points at an angle 50 relative to the x axis. Find the x and y components of this vector Y= r sin = (175m)(sin 50) = Answer 134 m

X = r cos (175m) (cos 50) = Answer 112 m

A jogger runs 145 m in a direction 20.0° east of north (displacement vector A) and then 105 m in a direction 35.0° south of east (displacement vector B). Using components, determine the magnitude and direction of the resultant vector C for these two displacements Ax= (145 m) sin 20.0°= 49.6 m Ay= (145 m) cos 20.0°= 136 m Bx = (105 m) cos 35.0° = 86.0 m By= -(105 m) sin 35.0° = 60.2 m Note that the component By is negative, because y points downward, in the negative y direction in the drawing. Substituting these values into the results for C and @ gives C= (Ax+Bx)^2+(Ay+By)^2=...