Answer:
Velocity will be [tex]v(t)=\frac{3t^2}{2}+5t-4ft/sec[/tex]
Distance traveled = 6 feet
Explanation:
We have given acceleration [tex]a(t)=3t+5[/tex]
We know that [tex]v(t)=\int a(t)dt[/tex]
So [tex]v(t)=\int 3t+5dt[/tex]
[tex]v(t)=\frac{3t^2}{2}+5t+c[/tex]
We have given [tex]v(0)=-4ft/sec[/tex]
So [tex]-4=\frac{3\times 0^2}{2}+5\times 0+c[/tex]
c = -4 ft/sec
So [tex]v(t)=\frac{3t^2}{2}+5t-4ft/sec[/tex]
Now we have to find distance traveled in interval [0,2]
So distance [tex]s=\int v(t)dt[/tex]
[tex]s=\int_{0}^{2}v(t)dt[/tex]
[tex]s=\int_{0}^{2}(\frac{3t^2}{2}+5t-4)dt[/tex]
[tex]s=(\frac{t^3}{2}+\frac{5t^2}{2}-4t)_{0}^{2}[/tex]
[tex]=(4+10-8)-0[/tex]
[tex]=6ft[/tex]
