A man is able to row a boat at 3 MPH in still water. If he rows his boat pointed straight across a river with a current of 4 MPH, what is his net velocity? If the river is 0. 5 miles wide, at what point will he land on the other side? Solution: The first step in problem solving is to identify the problem type. In this problem we are asked for a ‘net velocity. ‘ Since velocities behave as vectors, then we have a vector addition problem. A figure is drawn with the vectors indicated.

Here iv is the boat’s velocity still water), IVR is the river’s velocity, and event is the resultant (net) of these two velocities. The vector equation pertinent to the problem is: event = iv O IVR . iv IVR v net A coordinate system is selected as shown. Note that all vector problems are calculated by means of the component method (rectangular resolution), and hence a specified coordinate system is essential. Since our vector space is 2-dimensional, the vector equation above reduces to two scalar equations.

One for each of the necessary components. We calculate the components of the vectors to be added. For the coordinate system selected we have: event x = VBG + IVR = 3 + 3 MPH ; event y = by + vary = O + 4 = 4 MPH The answer for the net velocity can be specified in two equivalent ways. One way is to specify the two components for the net velocity (as done above). The second method is to state the magnitude and direction for event. Eleven = 3+42=25 = 5 MPH. The direction is given by an angle. From the figure we see we have a 3,4,5 right triangle.

Hence, we can determine B by any of the following: sin B = ‘ivy/event = 4/5 coos B = event = 3/5 tan B = = 4/3 = 530 = 530 = 530 3 4 1-2 The problem also asks at what point he will reach the other side of the river. Since all velocities in the problem are constant we may use: speed x time = distance. Thus the time to cross the river = 0. 5/3 = 1/6 = . 166 hrs = 10 min. Distance traveled downstream = = 2/3 mile. Total distance traveled = = 5/6 mile. He will arrive at a point on the other bank 2/3 off mile below his starting point. 2) A boy walks 4 km west, 2 km south, and then 1 km east.

Determine the total distance he walked and his net displacement. If it took him 1. 2 hours to complete the walk, find his average speed and his average velocity. Solution: We are dealing with ‘distances’ (scalar), ‘displacements’ (vector), ‘speeds’ (scalar), and Velocities’ (vectors). Hence we have both scalar addition and vector addition problems. We draw a figure indicating the 3 displacement vectors. Del B Dent DES We select a coordinate system (W & S). The total distance walked is simply 7 km (scalar addition). Since he covered this distance in 1. Hours, then his walking speed was: 7/1. 2 = 5. 83 km/hrs. Displacement is a vector quantity. Hence: Dent = Del DO 0 DO . The calculation is performed by adding the components of each vector. Thus: Dent W = DID + + DEW = 4+0- 1 = 3 km; Dent S = DISC + DES + DB’S = O + 2 2 km. Note that the West’ component of displacement DO is negative since the direction is east. The magnitude and direction can be determined from the resultant components. I Detent = 32+22 13 = 3. 61 km. B = 33. 70. In words, the total The direction can be found from any of the trig functions.

Tan B = DES/DO = 2/3 By definition, the average velocity is given by Arm where Ar is the total (net) displacement. Hence, eave = 3. 61/1. 2 = 3. 01 km/hrs at 33. 70 south of west. Note that the magnitude of the average velocity is not the same as the Walking speed. ‘ This illustrates the directional aspect of velocity. Suppose you walked for two hours and ended up back at your starting position. Your net displacement would be O, and your average velocity is also O. Note also the relationship between position and displacement.

The position vector is drawn from the selected origin of your CSS to the object’s location. Thus your ‘new position vector is the sum of the original position vector plus the displacement vector. For the 3 legs of the given trip we would have: 01-3 = or O Del (where or O by our CSS choice). RL e Danna 0 DB. Combining these we then have: or O Del O DO 0 DO . Since the original position as zero, the final position is identical to the net displacement. 3) Two helicopters take off from the same landing area. One proceeds 10 km at 500 south of west. The second flies 8 km in the direction 650 south of east.