Answer:
s= 6.53 m
t=3.27 s
Explanation:
velocity = 6 m/s
deceleration = -1.5[tex]v^\frac{1}{2}[/tex]
[tex]a=-1.5v^\frac{1}{2}\\v\frac{\mathrm{d} v}{\mathrm{d} s}=-1.5v^\frac{1}{2}\\-1.5ds=v^\frac{1}{2}dv\\\int\ {-1.5} \, ds= \int\ v^\frac{1}{2}dv\\-1.5s=\frac{2}{3}\times v^{\frac{3}{2} }[/tex]
now inserting value of v=6s we get distance(s)
s= 6.53 m ( distance cannot be negative)
now for time calculation we know that
[tex]a=\frac{\mathrm{d}v }{\mathrm{d} t}[/tex]
[tex]-1.5v^\frac{1}{2} =\frac{\mathrm{d}v }{\mathrm{d} t}\\-1.5dt=\frac{dv}{v^\frac{1}{2}} \\\int -1.5 dt=\int v^{-\frac{1}{2}}dt \\1.5t=2v^\frac{1}{2}\\t=\frac{4}{3}v^\frac{1}{2}[/tex]
putting value of v=6s
t=3.27 s (time cannot be negative)