Problem 6.25 Challenge

หาเวลาที่แรงมีค่าสูงสุดจาก \(\frac{dF(t)}{dt}= 0\)

\(\frac{d(at - bt^{2})}{dt}= 0\)

\(a - 2bt = 0\)

\(t =\frac{a}{2b}\)

หาแรงสูงสุดจาก \(F_{\max}= a(\frac{a}{2b}) - b(\frac{a}{2b})^{2} =\frac{a^{2}}{4b}\)

หาการดลจาก \(\vec{I} = \int \vec{F}dt\)

\(_{a/4b}\)

\(|\vec{I}|=|\int (at - bt^{2})dt|\)

\(_{0}\)

\(_{a/4b}\)

\(|\vec{I}| = |\frac{at^{2}}{2}-\frac{bt^{3}}{3}|=\frac{5a^{3}}{192b^{2}}\)

\(_{0}\)

หาแรงเฉลี่ยจาก \(\vec{I} = \vec{F}_{av}\Delta t\)

\(|\vec{F}_{av}| =\frac{|\vec{I}|}{\Delta t}=\frac{\frac{5a^{3}}{192b^{2}}}{\frac{a}{4b}}=\frac{5a^{2}}{48b}\)

\(\therefore |F_{\max}| - |\vec{F}_{av}| =\frac{a^{2}}{4b}-\frac{5a^{2}}{48b}=\frac{7a^{2}}{48b}\;\blacksquare\)