Respuesta :
Answer:
Step-by-step explanation:
Given
car start from rest at t=0
[tex]a=-0.6t+4 m/s^2[/tex]
[tex]\frac{\mathrm{d} v}{\mathrm{d} t}=a[/tex]
[tex]\frac{\mathrm{d} v}{\mathrm{d} t}=-0.6t+4[/tex]
[tex]dv=(-0.6t+4)dt[/tex]
integrating
[tex]\int dv=\int \left ( -0.6t+4\right )dt[/tex]
[tex]v=-0.3t^2+4t+c[/tex]
at [tex]t=0\ v=0[/tex]
thus [tex]c=0[/tex]
[tex]v=-0.3t^2+4t[/tex]
[tex]\frac{\mathrm{d} s}{\mathrm{d} t}=v[/tex]
[tex]ds=vdt[/tex]
integrating
[tex]\int ds=\int vdt[/tex]
[tex]\int ds=\int (-0.3t^2+4t)dt[/tex]
[tex]s=-0.1t^3+2t^2+c[/tex]
at [tex]t=0\, s=0[/tex]
thus [tex]c=0[/tex]
[tex]s=-0.1t^3+2t^2[/tex]
at [tex]s=100[/tex]
[tex]0.1t^3-2t^2+100=0[/tex]
Solving cubic equation
[tex]t=10\ s\ ,t=16.18\ s, t=-6.18\ s[/tex]
Neglecting negative value of t
we get [tex]t=10 s[/tex]
