# Physics Problem

01-1

1). 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 vb is the boat’s velocity (still water), vr is the river’s velocity, and vnet is the resultant (net) of these two velocities. The vector equation pertinent to the problem is: vnet = vb ⊕ vr .

vb vr

y

x

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: vnet x = vbx + vrx = 3 + 0 = 3 mph ; vnet y = vby + vry = 0 + 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 vnet. |vnet| =

3 + 4 2 = 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 θ by any of the following: sin θ = vy/vnet = 4/5 cos θ = vx/vnet = 3/5 tan θ = vy/vx = 4/3 θ = 530 θ = 530 θ = 530

3

θ

x

The equivalent answer for vnet is: 5 mph at 530 downstream.

4

y

v net

01-2 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 hr = 10 min. Distance traveled downstream = (1/6)(4) = 2/3 mile. Total distance traveled = (1/6)(5) = 5/6 mile. He will arrive at a point on the other bank 2/3 of a 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.

W D2

D1 θ Dnet D3 S

We select a coordinate system (W & S). The total distance walked is simply 7 km (scalar addition). Since he covered this distance in 1.2 hours, then his walking speed was: 7/1.2 = 5.83 km/hr. Displacement is a vector quantity. Hence: Dnet = D1 ⊕ D2 ⊕ D3 . The calculation is performed by adding the components of each vector. Thus: Dnet W = D1W + D2W + D3W = 4 + 0 – 1 = 3 km; Dnet S = D1S + D2S + D3S = 0 + 2 + 0 = 2 km. Note that the ‘West’ component of displacement D2 is negative since the direction is east. The magnitude and direction can be determined from the resultant components. |Dnet| =

32 + 2 2

=

13

= 3.61 km.

θ = 33.70. In words, the total

The direction can be found from any of the trig functions. Tan θ = DS/DW = 2/3 displacement is 3.61 km at 33.70 south of west.

By definition, the average velocity is given by Δr/Δt where Δr is the total (net) displacement. Hence, vave = 3.61/1.2 = 3.01 km/hr 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...

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