Hwk1_soln

Hwk1_soln - Phys 2101 Fall 2007 – Homework#1 –...

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Unformatted text preview: Phys 2101 Fall 2007 – Homework #1 – Solutions 1. The following graphs show position as a function of time for a particle that moves along the x axis. Select the graphs that at t = 1 s (the vertical hash mark across the x-axis is one second), fit the following situations. a. zero velocity and positive acceleration: [1], since the graph has a zero slope (velocity), increasing from negative to positive. Graph [2] has also zero velocity at t=1s, but it has negative acceleration. Graph [6] has positive acceleration, but not zero velocity at t=1s. b. zero velocity and negative acceleration: [2]. c. negative velocity and positive acceleration: [6] d. negative velocity and negative acceleration: [4] e. In which graph(s) would the speed of the particle be increasing and not zero at t = 1 s?: [4]. Notice that the velocity is decreasing because it is becoming more negative, but the speed (magnitude of the velocity, or steepness of the slope) is increasing. 2. How far does the runner whose velocity-time graph is shown below travel in 16 s? Since / v dx dt = (Eq. 2-4), then " x = v ( t ) dt # , which corresponds to the area under the v vs t graph. Dividing the total area A into rectangular (base × height) and triangular (1/2)(base x height) areas, we have " x = 1 2 (2 s )(8 m / s ) + (8 s )(8 m / s ) + (2 s )(4 m / s ) + 1 2 (2 s )(4 m / s ) + (4 s )(4 m / s ) = 100 m 3. The two vectors and in the figure below have equal magnitudes of 10.0 m and their angles are В 1 = 30° and В 2 = 105°. It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since r r a b , and r r form an isosceles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle r b makes with the + x axis is 30° +105° = 135° and apply Eq. 3-5 and Eq. 3- 6 where appropriate. (a) The x component of r r is r x = (10.0 m) cos 30° + (10.0 m) cos 135° = 1.59 m. (b) The y component of r r is r y = (10.0 m) sin 30° + (10.0 m) sin 135° = 12.1 m. (c) The magnitude of r r is 2 2 | | (1.59 m) (12.1 m) 12.2 m. r r = = + = r (d) The angle between r r and the + x direction is tan –1 [(12.1 m)/(1.59 m)] = 82.5°. 4. In the figure below, a vector a with a magnitude of 17.0 m is directed at an angle В = 56.0° counterclockwise from the +x axis. (a) With a = 17.0 m and θ = 56.0° we find a x = a cos θ = 9.51 m. (b) Similarly, a y = a sin θ = 14.1 m. (c) The angle relative to the new coordinate system is θ ´ = (56.0° – 18.0°) = 38.0°. Thus, ' cos ' 13.4 m. x a a ! = = (d) Similarly, ' y a = a sin θ ´ = 10.5 m. 5. Two vectors, r and s, lie in the xy plane. Their magnitudes are 4.50 and 7.30 units, respectively, and their directions are 320° and 85°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?...
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Hwk1_soln - Phys 2101 Fall 2007 – Homework#1 –...

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