motion in one dimension\n3. use the velocity - time graphs below to determine the acceleration. psyw\n4. the…

motion in one dimension\n3. use the velocity - time graphs below to determine the acceleration. psyw\n4. the area under the line of a velocity - time graph can be calculated using simple rectangle and triangle equations. the graphs below are examples:\nif the area under the line forms a...\nrectangle, then use\narea = base*height\ntriangle, then use\narea = 0.5*base*height\ntrapezoid, then make it into a rectangle + triangle and add the two areas.\na=(6 m/s)*(6 s)=36 m\na = 0.5*(6 m/s)*(6 s)=18 m\natotal = a rectangle+a triangle\natotal=(2m/s)*(6 s)+0.5*(4 m/s)*(6 s)=24 m\nfind the displacement of the objects represented by the following velocity - time graphs.\n5. for the following pos - time graphs, determine the corresponding shape of the vel - time graph.\n© the physics classroom, 2009\npage 16

motion in one dimension\n3. use the velocity - time graphs below to determine the acceleration. psyw\n4. the area under the line of a velocity - time graph can be calculated using simple rectangle and triangle equations. the graphs below are examples:\nif the area under the line forms a...\nrectangle, then use\narea = base*height\ntriangle, then use\narea = 0.5*base*height\ntrapezoid, then make it into a rectangle + triangle and add the two areas.\na=(6 m/s)*(6 s)=36 m\na = 0.5*(6 m/s)*(6 s)=18 m\natotal = a rectangle+a triangle\natotal=(2m/s)*(6 s)+0.5*(4 m/s)*(6 s)=24 m\nfind the displacement of the objects represented by the following velocity - time graphs.\n5. for the following pos - time graphs, determine the corresponding shape of the vel - time graph.\n© the physics classroom, 2009\npage 16

Answer

  1. For question 3 (finding acceleration from velocity - time graph):
    • Left - hand graph:

      • Explanation:

        Step1: Recall acceleration formula

        The acceleration (a=\frac{\Delta v}{\Delta t}), where (\Delta v = v_f - v_i) and (\Delta t=t_f - t_i). From the graph, (v_i = 5\ m/s), (v_f=30\ m/s), (t_i = 0\ s), (t_f = 8\ s). (\Delta v=30 - 5=25\ m/s), (\Delta t = 8-0 = 8\ s). (a=\frac{30 - 5}{8}=\frac{25}{8}=3.125\ m/s^{2})

        Step2: No further steps needed

        • Answer: (3.125\ m/s^{2})

    • Right - hand graph:

      • Explanation:

        Step1: Recall acceleration formula

        The acceleration (a=\frac{\Delta v}{\Delta t}), where (\Delta v = v_f - v_i) and (\Delta t=t_f - t_i). From the graph, (v_i = 30\ m/s), (v_f = 10\ m/s), (t_i=0\ s), (t_f = 12\ s). (\Delta v=10 - 30=- 20\ m/s), (\Delta t=12 - 0 = 12\ s). (a=\frac{10 - 30}{12}=-\frac{20}{12}=-\frac{5}{3}\approx - 1.67\ m/s^{2})

        Step2: No further steps needed

        • Answer: (-\frac{5}{3}\ m/s^{2}\approx - 1.67\ m/s^{2})

  2. For question 4 (finding displacement from velocity - time graph):
    • Left - hand graph:

      • Explanation:

        Step1: Identify shape and formula

        The graph is a rectangle. The formula for the area of a rectangle is (A = base\times height). Here, the base (time) (t = 8\ s) and the height (velocity) (v = 8\ m/s). (A=(8\ m/s)\times(8\ s)=64\ m)

        Step2: No further steps needed

        • Answer: (64\ m)

    • Middle graph:

      • Explanation:

        Step1: Identify shape and formula

        The graph is a triangle. The formula for the area of a triangle is (A = 0.5\times base\times height). Here, the base (time) (t = 8\ s) and the height (velocity) (v = 12\ m/s). (A = 0.5\times(8\ s)\times(12\ m/s)=48\ m)

        Step2: No further steps needed

        • Answer: (48\ m)

    • Right - hand graph:

      • Explanation:

        Step1: Split into rectangle and triangle

        Split the shape into a rectangle and a triangle. For the rectangle, base (t_1=4\ s), height (v_1 = 4\ m/s), so (A_{rectangle}=(4\ m/s)\times(4\ s)=16\ m). For the triangle, base (t_2 = 8\ s), height (v_2=12 - 4=8\ m/s), so (A_{triangle}=0.5\times(8\ s)\times(8\ m/s)=32\ m).

        Step2: Calculate total area

        (A_{total}=A_{rectangle}+A_{triangle}=16 + 32=48\ m)

        • Answer: (48\ m)