Question

A marble slides along a frictionless track at a constant speed u before encountering a section of the track that is covered with a thin layer of putty. The instantaneous velocity v of the marble as it slides past the putty, as a function of time t, is given byv(t)=u-bt+ct^3where t = 0 at the moment when the marble first comes into contact with the putty. As the marble exits the region covered in putty, the marble\'s instantaneous acceleration goes to zero. For u = 2.50 m/s, b = 3.55 m/s2, and c = 6.70 m/s4, calculate the length of the track that is covered with putty._______m

Answer 1

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

$v(t)=u-bt+ct^3$

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.

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(u-bt+ct^3)}{dt}\\\\a=-b+3ct^2$

or

$0-3.55+3\times 6.7t^2\\\\0=-3.55+20.1t^2\\\\t=\sqrt{\dfrac{3.55}{20.1}} \\\\t=0.42\ s$

Let x is the length of the track.

$v=\dfrac{dx}{dt}\\\\\text{or}\\\\\int\limits_0^x {dx} =\int\limits^{4.2}_0 {vdt} \\\\x=\int\limits^{4.2}_0 {u-bt+ct^3} \, dt\\\\x=[ut-\dfrac{bt^2}{2}+\dfrac{ct^4}{4}]_0^ {0.42}\\\\\text{Putting limits}\\\\x=0.42u-\dfrac{b(0.42)^2}{2}+\dfrac{c(0.42)^4}{4}\\\\\text{Now, put values of u,b and c}\\\\x=0.42(2.5)-\dfrac{3.55\times (0.42)^2}{2}+\dfrac{6.7\times (0.42)^4}{4}\\\\x=0.789\ m\\\\x=78.9\ cm$

So, the length of the track that is covered with putty is 78.9 cm.