Question

A motorist drives south at 28.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.40 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.40 min trip, find the following values. Let the positive x axis point east.

(a) total vector displacement
m (magnitude) ° south of west

(b) average speed
m/s

(c) average velocity
m/s (magnitude) ° south of west

Answer 1

Answer on Question #39202, Physics, Other

A motorist drives south at $28.0\mathrm{m / s}$ for $3.00\mathrm{min}$ , then turns west and travels at $25.0\mathrm{m / s}$ for $2.40\mathrm{min}$ , and finally travels northwest at $30.0\mathrm{m / s}$ for $1.00\mathrm{min}$ . For this $6.40\mathrm{min}$ trip, find the following values. Let the positive $x$ axis point east.

(a) total vector displacement M magnitude. deg south of west

(b) average speed m/s

(c) average velocity $\mathrm{m / s}$ (magnitude) $\mathrm{{}^\circ}$ south of west

Solution:

(a) Let construct three vectors and find their sum:

Distance=velocity x time

$$\overline {{d 1}} = - 2 8 * 3 * 6 0 \vec {j} = - 5 0 4 0 \vec {j}$$ $$\overline {{d 2}} = - 2 5 * 2. 4 * 6 0 \vec {i} = - 3 6 0 0 \vec {i}$$ $$\overline {{d 3}} = - 3 0 \cos 4 5 {}^ {\circ} * 6 0 \vec {i} + 3 0 \sin 4 5 {}^ {\circ} * 6 0 \vec {j} = - 1 2 7 3 \vec {i} + 1 2 7 3 \vec {j}$$

Total vector displacement:

$$\vec {d} = \vec {d 1} + \vec {d 2} + \vec {d 3} = - 5 0 4 0 \vec {j} - 3 6 0 0 \vec {i} - 1 2 7 3 \vec {i} + 1 2 7 3 \vec {j} = - 4 8 7 3 \vec {i} - 3 7 6 7 \vec {j}$$

The magnitude of the displacement is

$$d = \sqrt {(x) ^ {2} + (y) ^ {2}} = \sqrt {(4 8 7 3) ^ {2} + (3 7 6 7) ^ {2}} = 6 1 6 0 m$$

The direction of vector $\mathbf{d}$ :

$$\tan \theta = \frac {y}{x} = \frac {3 7 6 7}{4 8 7 3}$$ $$\theta = \arctan (0. 7 7 3) = 3 7. 7 {}^ {\circ}$$

(b) The average speed of an object in an interval of time is the distance travelled by the object divided by the duration of the interval.

$$\bar {v} = \frac {d 1 + d 2 + d 3}{t 1 + t 2 + t 3} = \frac {5 0 4 0 + 3 6 0 0 + 1 8 0 0}{(3 + 2 . 4 + 1) * 6 0} = 2 7. 1 8 7 5 \approx 2 7. 1 9 \mathrm {m / s}$$

(c) The magnitude of an average velocity

$$v = \frac {d}{t 1 + t 2 + t 3} = \frac {6 3 8 5 . 8}{(3 + 2 . 4 + 1) * 6 0} \approx 1 6. 6 3 \mathrm {m / s}$$

The direction of an average velocity 40.26° south of west.

**Answer.**

(a) total vector displacement = -4873i - 3767j; magnitude 6160 m; 37.7° south of west

(b) average speed 27.19 m/s

(c) average velocity 16.63 m/s (magnitude) 40.26° south of west