Physics Midterm Study Guide
The eye of a hurricane passes over New Orleans in a direction 60.0° north of west with a speed of 41.0 km/hr. Three hours later, the hurricane suddenly shifts due north, and its speed slows to 25.0 km/hr. How far from New Orleans is the hurricane 4.50 hours after it passes over the city? v1 = 41.0km/hr(cos30°, sin30°) = (35.5, 20.5) km/hr 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
→ 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