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} )

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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)

Answer

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)

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 \( 5 \mathrm{~s} \)

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