Hifumi's Study NotesnotesCalculusCegep2Rectilinear Motion

Rectilinear Motion

Tags

Calculus

Cegep/2

Word count

318 words

Reading time

2 minutes

Motion in a straight line

Displacement:

\int_{t_{1}}^{t_{2}} v (t) d t = s (t_{2}) - s (t_{1})

Distance travelled:

\int_{t_{1}}^{t_{2}} | v (t) | d t

Examples

An object moves along the x-axis with an acceleration given by $a (t) = \cos t - \sin t$ with initial velocity of $1 \frac{cm}{s}$ .
Find both the displacement and the distance travelled from $t_{1} = 0$ to $t_{2} = π$ .
If initial position is $3 cm$ , find the position function $s (t)$ .

\begin{aligned} v (t) & = \int a (t) d t \\ = \int (\cos t - \sin t) d t \\ = \sin t + \cos t + C \\ 1 & = \sin 0 + \cos 0 + C \\ C & = 0 \\ ∴ v (t) & = \sin t + \cos t \\ displacement & = \int_{t_{1}}^{t_{2}} v (t) d t \\ = \int_{0}^{π} (\sin t + \cos t) d t \\ = [- \cos t + \sin t]_{0}^{π} \\ = - \cos π + \sin π - (- \cos 0 + \sin 0) \\ = 2 cm \end{aligned}

We need to determine the sign of $v (t)$ .

\begin{aligned} \sin t + \cos t & = 0 \\ \sin t & = - \cos t \\ \tan t & = - 1 \\ t & = \frac{3 π}{4} \\ distance travelled & = \int_{0}^{π} | \sin t + \cos t | d t \\ = - \int_{0}^{\frac{3 π}{4}} (\sin t + \cos t) d t + \int_{\frac{3 π}{4}}^{π} (\sin t + \cos t) d t \\ = - (- \cos t + \sin t)_{0}^{\frac{3 π}{4}} + (- \cos t + \sin t)_{\frac{3 π}{4}}^{π} \\ = - (- \cos \frac{3 π}{4} + \sin \frac{3 π}{4}) - (- \cos 0 + \sin 0) - \cos π + \sin π - (- \cos \frac{3 π}{4} + \sin \frac{3 π}{4}) \\ = 2 \sqrt{2} cm \end{aligned}

\begin{aligned} s (t) & = \int v (t) d t \\ = \int v (t) d t \\ = \int (\sin t + \cos t) d t \\ = - \cos t + \sin t + C_{2} \\ 3 & = - \cos 0 + \sin 0 + C_{2} \\ C_{2} & = 4 \\ ∴ s (t) & = - \cos t + \sin t + 4 \end{aligned}

Contributors

Changelog

Last edited about 2 hours ago

View full history