Answer:
Part a : Time required for the block to go to 3.05 in is 0.036 s
Part b : Time required for the block to go to 6.1 in is 0.054 s
Explanation:
The acceleration is given as
[tex]a=a_o-kx[/tex]
Here
[tex]k=\frac{a_o}{l_c-l_{uc}}[/tex]
[tex]k=\frac{420}{\frac{16-9.9}{12}}\\k=826.22 s^{-2}[/tex]
So now the equation becomes
[tex]a=420-826.23x[/tex]
As per relation of acceleration to velocity
[tex]vdv=adx[/tex]
[tex]\int\limits^v_0 {v} \, dv =\int a dx[/tex]
[tex]\frac{v^2}{2} =\int (420-826x) dx[/tex]
[tex]\frac{v^2}{2} =420x-413x^2[/tex]
[tex]{v^2}=840x-826x^2\\{v}=\sqrt{840x-826x^2}\\[/tex]
Also
[tex]v=\frac{dx}{dt}[/tex]
So
[tex]\frac{dx}{dt}=\sqrt{840x-826x^2}\\[/tex]
Solving for dt and integrating results in
[tex]{dt}=\frac{dx}{\sqrt{840x-826x^2}}\\\int\limits^t_0 \,dt=\int\limits^x_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=\int\limits^x_0 \,\frac{dx}{\sqrt{840x-826x^2}}[/tex]
Now using this equation value for both parts can be calculated numerically as
Part a 3.05 in
[tex]x=\frac{3.05}{12}\\x=0.254 ft[/tex]
Solving numerically for this value of x
[tex]t=\int\limits^{0.254}_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=0.036 s[/tex]
Time required for the block to go to 3.05 in is 0.036 s
Part b 6.1 in
[tex]x=\frac{6.1}{12}\\x=0.504 ft[/tex]
Solving for this value of x
[tex]t=\int\limits^{0.504}_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=0.0543 s[/tex]
Time required for the block to go to 6.1 in is 0.054 s
