Answer to Question #6047 in Mechanics | Relativity for Terence
Question #6047
From t= 0 s to t= 13 s, a car moves east with a constant speed of 11 m/s. Then, from t = 13 s to t = 32 s, the car moves west with a constant speed of 14 m/s. Lastly, from t = 32 s to t = 51 s, the car moves west with a constant speed of 26 m/s. Determine the total distance the car traveled, the car's total displacement, it's average speed, & it's average velocity. Express all answers in the appropriate mks units.
A.) Total distance traveled = 1
B.) Total displacement = 2 3 --- Choose direction --- north south east west zero vector
C.) Average speed = 4
D.) Average velocity = 5 6 --- Choose direction --- north south east west zero vector
A.) Total distance traveled = 1
B.) Total displacement = 2 3 --- Choose direction --- north south east west zero vector
C.) Average speed = 4
D.) Average velocity = 5 6 --- Choose direction --- north south east west zero vector
Expert's answer
Movement at the east direction:
Le = 11*(13-0) = 143 m.
Movement at the west direction:
Lw = 14*(32-13) = 266 m.
East and west are directed oppositely, so displacement is
D = 266 - 143 = 123 m west.
Movement at the west direction:
Lw = 26*(51-32) = 494.
So, the total displacement is
D = 494+123 = 617 m west.
The total travelled distance is
L = 143+266+123+494 = 1026 m.
Average car's speed (velocity) is
V =& L/T = 1026m/51s ≈ 20.1176 m/s.
Le = 11*(13-0) = 143 m.
Movement at the west direction:
Lw = 14*(32-13) = 266 m.
East and west are directed oppositely, so displacement is
D = 266 - 143 = 123 m west.
Movement at the west direction:
Lw = 26*(51-32) = 494.
So, the total displacement is
D = 494+123 = 617 m west.
The total travelled distance is
L = 143+266+123+494 = 1026 m.
Average car's speed (velocity) is
V =& L/T = 1026m/51s ≈ 20.1176 m/s.
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