1. A student drives south at 80m/s for four minutes, then turns west and drives at 100m/s for five minutes, and finally travels northwest at 120m/s for three minute. For this twelve minutes trip, find (a)the total vector displacement, (b) the average speed, and (c) the average velocity.

To solve this problem, we need to break down the trip into its components and calculate each part separately.<br /><br />### (a) Total Vector Displacement<br /><br />1. **Southward Displacement:**<br /> - Speed = 80 m/s<br /> - Time = 4 minutes = 240 seconds<br /> - Displacement = Speed × Time = \(80 \, \text{m/s} \times 240 \, \text{s} = 19200 \, \text{m}\) south<br /><br />2. **Westward Displacement:**<br /> - Speed = 100 m/s<br /> - Time = 5 minutes = 300 seconds<br /> - Displacement = Speed × Time = \(100 \, \text{m/s} \times 300 \, \text{s} = 30000 \, \text{m}\) west<br /><br />3. **Northwestward Displacement:**<br /> - Speed = 120 m/s<br /> - Time = 3 minutes = 180 seconds<br /> - Displacement = Speed × Time = \(120 \, \text{m/s} \times 180 \, \text{s} = 21600 \, \text{m}\) northwest<br /><br /> Since northwest is at a 45-degree angle, we can split this into north and west components:<br /> - North component = \(21600 \, \text{m} \times \cos(45^\circ) = 15273.6 \, \text{m}\)<br /> - West component = \(21600 \, \text{m} \times \sin(45^\circ) = 15273.6 \, \text{m}\)<br /><br />**Total Displacement Components:**<br />- South-North Component: \(19200 \, \text{m} \, \text{south} - 15273.6 \, \text{m} \, \text{north} = 3926.4 \, \text{m} \, \text{south}\)<br />- West Component: \(30000 \, \text{m} + 15273.6 \, \text{m} = 45273.6 \, \text{m} \, \text{west}\)<br /><br />**Resultant Displacement:**<br />Using Pythagorean theorem:<br />\[ \text{Magnitude} = \sqrt{(3926.4)^2 + (45273.6)^2} \approx 45443.8 \, \text{m} \]<br /><br />The direction can be found using:<br />\[ \theta = \tan^{-1}\left(\frac{3926.4}{45273.6}\right) \approx 4.96^\circ \, \text{south of west} \]<br /><br />### (b) Average Speed<br /><br />Average speed is the total distance traveled divided by the total time.<br /><br />- Total distance = \(19200 \, \text{m} + 30000 \, \text{m} + 21600 \, \text{m} = 70800 \, \text{m}\)<br />- Total time = 12 minutes = 720 seconds<br /><br />\[ \text{Average Speed} = \frac{70800 \, \text{m}}{720 \, \text{s}} = 98.33 \, \text{m/s} \]<br /><br />### (c) Average Velocity<br /><br />Average velocity is the total displacement divided by the total time.<br /><br />\[ \text{Average Velocity Magnitude} = \frac{45443.8 \, \text{m}}{720 \, \text{s}} \approx 63.11 \, \text{m/s} \]<br /><br />The direction is the same as the total vector displacement, approximately \(4.96^\circ\) south of west.