Respuesta :
Answer:
x = 78.9 cm
Explanation:
The instantaneous velocity v of the marble as it slides past the putty, as a function of time t, is given by
[tex]v(t)=u-bt+ct^3[/tex]
u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴
As the marble exits the region covered in putty, the marble\'s instantaneous acceleration goes to zero.
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(u-bt+ct^3)}{dt}\\\\a=-b+3ct^2[/tex]
or
[tex]0-3.55+3\times 6.7t^2\\\\0=-3.55+20.1t^2\\\\t=\sqrt{\dfrac{3.55}{20.1}} \\\\t=0.42\ s[/tex]
Let x is the length of the track.
[tex]v=\dfrac{dx}{dt}\\\\\text{or}\\\\\int\limits_0^x {dx} =\int\limits^{4.2}_0 {vdt} \\\\x=\int\limits^{4.2}_0 {u-bt+ct^3} \, dt\\\\x=[ut-\dfrac{bt^2}{2}+\dfrac{ct^4}{4}]_0^ {0.42}\\\\\text{Putting limits}\\\\x=0.42u-\dfrac{b(0.42)^2}{2}+\dfrac{c(0.42)^4}{4}\\\\\text{Now, put values of u,b and c}\\\\x=0.42(2.5)-\dfrac{3.55\times (0.42)^2}{2}+\dfrac{6.7\times (0.42)^4}{4}\\\\x=0.789\ m\\\\x=78.9\ cm[/tex]
So, the length of the track that is covered with putty is 78.9 cm.
The length of the track is covered with putty is 0.789 m.
How do you calculate the length of the track?
The instantaneous velocity v of the marble as a function of time t is given below.
[tex]v(t)=u-bt+ct^3[/tex]
Where t = 0 at the moment when the marble first comes into contact with the putty. Also given that, u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴.
Given that the acceleration = 0 as the marble exits the region covered in putty. The acceleration is the rate change in the velocity with respect to time.
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]a = \dfrac {d(u-bt+ct^3)}{dt}[/tex]
[tex]a = -b + 3ct^2[/tex]
Substituting the values in the above equation, we get the time.
[tex]0 = -3.55 + 3 \times 6.70 t^2[/tex]
[tex]t^2 = 0.1766[/tex]
[tex]t = 0.42 \;\rm s[/tex]
Hence the time interval is t =0 to t = 0.42 seconds.
Let us consider that the length of the track is l. Then the velocity can be written as the function of length is given below.
[tex]v = \dfrac {dl}{dt}[/tex]
[tex]dl = vdt[/tex]
[tex]\int_{0}^{t}dl = \int_{0}^{0.42} vdt[/tex]
[tex]l = \int_{0}^{0.42} (u-bt+ct^3) dt[/tex]
[tex]l =[ ut-\dfrac {bt^2}{2} + \dfrac {ct^4}{4} ]_0^{0.42}[/tex]
Putting the value t = 0.42,
[tex]l = 2.50 \times 0.42 - \dfrac {3.55 \times (0.42)^2}{2} + \dfrac { 6.70 \times (0.42)^4}{4}[/tex]
[tex]l = 0.789 \;\rm m[/tex]
Hence the length of the track is covered with putty is 0.789 m.
To know more about velocity and acceleration, follow the link given below.
https://brainly.com/question/2437624.
