The given position function is:
[tex]s(t)=t^3-5t^2+3t[/tex]
The derivative of the position function gives the velocity function:
[tex]v(t)=s^{\prime}(t)[/tex]
Find the derivative of the position vector:
[tex]\begin{gathered} s^{\prime}(t)=3t^{3-1}-2\cdot5t^{2-1}+3t^{1-1} \\ \Rightarrow s^{\prime}(t)=3t^2-10t+3 \end{gathered}[/tex]
It follows that:
[tex]v(t)=3t^2-10t+3[/tex]
The derivative of the velocity function is the acceleration function:
[tex]a(t)=v^{\prime}(t)[/tex]
Find the derivative of the velocity function:
[tex]\begin{gathered} v^{\prime}(t)=3\cdot2t^{2-1}-10t^{1-1}+0 \\ \Rightarrow v^{\prime}(t)=6t-10 \end{gathered}[/tex]
It implies that the acceleration is:
[tex]a(t)=6t-10[/tex]
Substitute t=1 into the acceleration function:
[tex]a(1)=6(1)-10=6-10=-4\text{ m/s}^2[/tex]
Hence, the acceleration at t=1 is -4m/s². ( the negative sign denotes a decrease in velocity)
