3. use the velocity - time graphs below to determine the acceleration. psyw\n4. the area under the line of a…

3. 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 ...\n... rectangle, then use area = base*height\n... triangle, then use area = 0.5 * base*height\n... trapezoid, 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+atriangle\natotal=(2 m/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.

3. 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 ...\n... rectangle, then use area = base*height\n... triangle, then use area = 0.5 * base*height\n... trapezoid, 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+atriangle\natotal=(2 m/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.

Answer

  1. For the problem of finding acceleration from velocity - time graphs (Question 3):
    • Explanation:
      • Step 1: Recall the formula for acceleration
        • Acceleration (a=\frac{\Delta v}{\Delta t}), where (\Delta v = v_f - v_i) (final velocity - initial velocity) and (\Delta t=t_f - t_i) (final time - initial time).
      • For the first velocity - time graph in Question 3:
        • (v_i = 5\ m/s), (v_f=30\ m/s), (t_i = 0\ s), (t_f = 8\ s).
        • (a=\frac{v_f - v_i}{t_f - t_i}=\frac{30 - 5}{8-0}=\frac{25}{8}=3.125\ m/s^{2}).
      • For the second velocity - time graph in Question 3:
        • (v_i = 30\ m/s), (v_f = 5\ m/s), (t_i=0\ s), (t_f = 12\ s).
        • (a=\frac{v_f - v_i}{t_f - t_i}=\frac{5 - 30}{12 - 0}=\frac{- 25}{12}\approx - 2.083\ m/s^{2}).
    • Answer: For the first graph, the acceleration is (3.125\ m/s^{2}). For the second graph, the acceleration is (-2.083\ m/s^{2}).
  2. For the problem of finding displacement from velocity - time graphs (Question 4):
    • Explanation:
      • For the first velocity - time graph in Question 4:
        • The graph is a rectangle with (v = 12\ m/s) and (t = 8\ s). Using the formula for the area of a rectangle (A=v\times t).
        • (A=(12\ m/s)\times(8\ s)=96\ m).
      • For the second velocity - time graph in Question 4:
        • The graph is a triangle with (v_i = 0\ m/s), (v_f = 12\ m/s) and (t = 8\ s). Using the formula for the area of a triangle (A = 0.5\times v\times t).
        • (A=0.5\times(12\ m/s)\times(8\ s)=48\ m).
      • For the third velocity - time graph in Question 4:
        • The graph can be split into a rectangle and a triangle. The rectangle has (v_{rect}=4\ m/s) and (t = 8\ s), and the triangle has (v_{triangle}=12 - 4=8\ m/s) and (t = 8\ s).
        • (A_{rect}=(4\ m/s)\times(8\ s)=32\ m), (A_{triangle}=0.5\times(8\ m/s)\times(8\ s)=32\ m).
        • (A_{total}=A_{rect}+A_{triangle}=32 + 32=64\ m).
    • Answer: For the first graph, the displacement is (96\ m). For the second graph, the displacement is (48\ m). For the third graph, the displacement is (64\ m).