Respuesta :
Answer:
Explanation:
Given
At [tex]t=0[/tex] velocity [tex]v_{0x}=20\ m/s[/tex]
at this instant acceleration [tex]a_x=-Ct[/tex]
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=-Ct[/tex]
[tex]\int_{20}^{0}dv=\int_{0}^{8}-Ctdt[/tex]
[tex]\int_{0}^{20}dv=\int_{0}^{8}Ctdt[/tex]
[tex]C=\frac{5}{8}[/tex]
(b)Distance traveled in 8 s
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=-Ct[/tex]
[tex]\int_{20}^{v}dv=\int_{0}^{t}-Ctdt[/tex]
[tex]v-20=-\frac{Ct^2}{2}[/tex]
[tex]v=20-\frac{Ct^2}{2}[/tex]
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\int_{0}^{8}vdt=\int_{0}^{x}dx[/tex]
[tex]\int_{0}^{8}20-\frac{Ct^2}{2}dt=\int_{0}^{x}dx[/tex]
[tex]x=20t|_{0}^{8}-\frac{C}{6}t^3|_{0}^{8}[/tex]
here value of [tex]C=\frac{5}{8}[/tex]
[tex]x=20\times 8-\frac{0.625}{6}\times 8^3[/tex]
[tex]x=106.67\ m[/tex]
The value of C if the object stops in 8.00 s is 0.625 m/s³.
The distance traveled by the object before stopping in 8 seconds is 40 m.
The given parameters;
- initial velocity, [tex]v_0[/tex] = 20.0 m/s
- initial time of motion, t = 0
- acceleration of the object, a = -Ct
The value of C is determined by using velocity equation as shown below;
[tex]\frac{dv}{dt} = -Ct\\\\dv = -Ctdt\\\\\int\limits^v_{v_0} \, dv = -\int\limits^t_{t_0} \, Ct \\\\v-v_0= -C[\frac{t^2}{2} ]^t_0\\\\v-v_0 = - \frac{1}{2} Ct^2\\\\0 - 20 = - \frac{1}{2}C(8)^2\\\\-20 = -32 C\\\\C = \frac{20}{32} = 0.625 \ m/s^3[/tex]
The acceleration of the object during 8 seconds is calculated as follows;
a = -Ct
a = -0.625(8)
a = -5 m/s²
The distance traveled by the object before stopping in 8 seconds is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + 2(-5)s\\\\0 = 400 - 10s\\\\10s = 400 \\\\s = \frac{400}{10} \\\\s = 40 \ m[/tex]
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