Answer:
1) Velocity at t =6 =32m/s
2) Position of particle at t = 6 secs = 67 meters
3) Distance covered in 6 seconds equals 72 meters.
Explanation:
By definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\dv=a(t)dt\\\\\int dv=\int a(t)dt\\\\v(t)=\int (2t-1)dt\\\\v(t)=t^{2}-t+c_{1}[/tex]
Now at t = 0 v = 2 m/s thus the value of constant is obtained as
[tex]2=0-0+c_{1}\\\\\therefore c_{1}=2[/tex]
thus velocity as a function of time is given by
[tex]v(t)=t^{2}-t+2[/tex]
Similarly position can be found by
[tex]x(t)=\int v(t)dt\\\\x(t)=\int (t^{2}-t+2)dt\\\\x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+c_{2}[/tex]
The value of constant can be obtained by noting that at time t = 0 x = 1.
Thus we get
[tex]1=0+0+0+c_{2}\\\\\therefore c_{2}=1[/tex]
thus position as a function of time is given by
[tex]x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+1[/tex]
Thus at t = 6 seconds we have
[tex]v(6)=6^{2}-6+2=32m/s[/tex]
[tex]x(6)=\frac{6^{3}}{3}-\frac{6^{2}}{2}+2\times 6 +1=67m[/tex]
The path length can be obtained by evaluating the integral
[tex]s=\int_{0}^{6}\sqrt{1+(t^{2}-t+2)^{2}}\cdot dt\\\\s=72meters[/tex]
