The position of a particle moving along \( x \) axis is given by \( x=1 t^{3}+8 t^{2}+35 \), find the instantaneous acceleration at \( 5 \mathrm{~s} \)

Step 1 ：\(x(t) = t^3 + 8t^2 + 35\)

Step 2 ：\(v(t) = \frac{dx}{dt} = 3t^2 + 16t\)

Step 3 ：\(a(t) = \frac{dv}{dt} = 6t + 16\)

Step 4 ：\(a(5) = 6(5) + 16\)

Step 5 ：\(a(5) = 46\)