Respuesta :
Answer and Solution:
As per the question:
The particle's displacement is given by the expression:
[tex]s = t^{2} - 5t + 16[/tex]
Now,
(a) The average velocity is given as the ratio of the displacement of the particle to the time interval:
[tex]v_{avg} = \frac{\Delta s}{\Delta t}[/tex]
Now, displacement in the interval:
i) [3,4]
At t = 3 s
[tex]s_{3} = 3^{2} - 5(3) + 16 = 10 m[/tex]
At t = 4 s
[tex]s_{4} = 4^{2} - 5(4) + 16 = 12 m[/tex]
[tex]v_{avg} = \frac{s_{4} - s_{3}}{4 - 3} = \frac{12 - 10}{4 - 3} = 2 m/s[/tex]
ii) [3.5, 4]
At t = 3.5 s
[tex]s_{3.5} = 3.5^{2} - 5(3.5) + 16 = 10.75 m[/tex]
At t = 4 s
[tex]s_{4} = 4^{2} - 5(4) + 16 = 12 m[/tex]
[tex]v_{avg} = \frac{s_{4} - s_{3.5}}{4 - 3.5} = \frac{12 - 10.75}{0.5} = 2.5 m/s[/tex]
iii) [4, 5]
At t = 4 s
[tex]s_{4} = 4^{2} - 5(4) + 16 = 12 m[/tex]
At t = 5 s
[tex]s_{5} = 5^{2} - 5(5) + 16 = 16 m[/tex]
[tex]v_{avg} = \frac{s_{5} - s_{4}}{5 - 4} = \frac{16 - 12}{1} = 4 m/s[/tex]
iv) [4, 4.5]
At t = 4 s
[tex]s_{4} = 4^{2} - 5(4) + 16 = 12 m[/tex]
At t = 4.5 s
[tex]s_{4.5} = 4.5^{2} - 5(4.5) + 16 = 13.75 m[/tex]
[tex]v_{avg} = \frac{s_{4.5} - s_{4}}{4.5 - 4} = \frac{13.75 - 12}{0.5} = 3.5 m/s[/tex]
(b) The instantaneous velocity at t = 4 s is given by:
[tex]v_{i} = \frac{ds}{dt}[/tex]
[tex]v_{i} = \frac{d}{dt}(t^{2} - 5t + 16) = 2t - 5[/tex]
At t = 4 s
[tex]v_{i} = 2(4) - 5 = 3\ m/s[/tex]
Answer:
Explanation:
Given
displacement [tex]s=t^2-5t+16[/tex]
average velocity is [tex]\frac{\int_{t_1}^{t_2}vdt}{\int_{t_1}^{t_2}dt}[/tex]
[tex]v_{avg}=\frac{\int_{t_1}^{t_2}(t^2-5t+16)dt}{\int_{t_1}^{t_2}dt}[/tex]
(i)t=3 to 4 s
[tex]v_{avg}=\frac{\left [ \frac{t^3}{3}-\frac{-5t^2}{2}+16\right ]_3^4}{\left [ t\right ]_3^4}[/tex]
[tex]v_{avg}=10.83 m/s[/tex]
(ii)t=3.5 s to 4 s
s at t=4 s
s=20 m
s at t=3.5 s, s=10.75 m
[tex]v_{avg}=\frac{20-10.75}{4-3.5}=18.5 m/s[/tex]
(iii)s at t=5 s
s=16 m
[tex]v_{avg}=\frac{16-20}{1}=-4 m/s[/tex]
velocity changes its direction
(iv) [4, 4.5 ]
s at t=4.5 s
s=13.75 m
[tex]v_{avg}=\frac{13.75-20}{0.5}=-12.5 m/s[/tex]
(b)instantaneous velocity[tex]=\frac{\mathrm{d} s}{\mathrm{d} t}=2t-5[/tex]
t=4
v=8-5=3 m/s
