Question

a radio-operated car travels 12.0 meters east in 4 seconds then 5.0 meters northin 2 seconds (yes, the car can make 90 degree turns without pausing). (A) Drawa vector map that describes the car's motion, including the car's displacement.(B) Calculate the car's average speed for the entire 6-second trip. (C) Calculatethe car's average velocity (magnitude and direction) for the 6-second trip.

Answer 1

Answer on Question# 38215, Physics, Other

A radio-operated car travels 12.0 meters east in 4 seconds then 5.0 meters north in 2 seconds (yes, the car can make 90 degree turns without pausing). (A) Draw a vector map that describes the car's motion, including the car's displacement.(B) Calculate the car's average speed for the entire 6-second trip. (C) Calculate the car's average velocity (magnitude and direction) for the 6-second trip.

Solution

(A)

$\vec{d}$ - the car's displacement.

(B) The car's average speed is

$$r = \frac {d}{t},$$

where $d$ - the total distance car travelled, $t$ - the total time of the 6-second trip.

$$r = \frac {1 2 m + 5 m}{4 s + 2 s} = 2. 8 \frac {m}{s}.$$

(C) The car's average velocity is

$$\vec {v} = \frac {\vec {s}}{t},$$

where $\vec{s}$ - the car's displacement, $t$ - the total time of the 6-second trip.

Magnitude of average velocity is

$$v = \frac {s}{t} = \frac {\sqrt {(1 2 m) ^ {2} + (5 m) ^ {2}}}{4 s + 2 s} = 2. 2 \frac {m}{s}.$$

Direction of average velocity is

$$\theta = \tan^ {- 1} \left(\frac {5}{1 2}\right) = 2 1. 8 {}^ {\circ} \text {north of east}.$$