Answer:
(a) No, the acceleration is not constant
(b) Yes, it is constant between 6 s and 12 s
Explanation:
Given;
time of motion; t(s) = 0 2 4 6 8 10 12 14 16
velocity (m/s); v = 0 0 2 5 10 15 20 22 22
Average acceleration for t= (2,0) and v = (0,0)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{0-0}{2-0} = 0 \ m/s^2[/tex]
Average acceleration for t = (4, 2) and v = (2,0)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{2-0}{4-2} = 1.0 \ m/s^2[/tex]
Average acceleration for t = (6, 4) and v = (5,2)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{5-2}{6-4} = 1.5 \ m/s^2[/tex]
Average acceleration for t = (8, 6) and v = (10,5)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{10-5}{8-6} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (10, 8) and v = (15,10)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{15-10}{10-8} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (12,10) and v = (20,15)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{20-15}{12-10} = 2.5 \ m/s^2[/tex]
Average acceleration for t = (14,12) and v = (22,20)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{22-20}{14-12} = 1.0 \ m/s^2[/tex]
Average acceleration for t = (22, 22) and v = (16,14)
[tex]a = \frac{v_2-v_1}{t_2-t_1} \\\\a = \frac{22-22}{16-14} = 0 \ m/s^2[/tex]
(a) B. No (The acceleration is not constant)
(b) A. Yes (it is constant between 6 s and 12 s)
