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Question

# A particle P moves in a straight line with position given by $s(t)=80 e^{-\frac{t}{10}}-40 t \mathrm{m}$ where t is the time in seconds, $t \geqslant 0$. Find the initial position, velocity, and acceleration of P.

Solution

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### Plug in $t=0$ to find the initial conditions:

\begin{aligned} s(0) &= 80 e^{-0/10} - 40(0) = 80 \qquad \text{ m} \\\\ v(0) &= -8 e^{-0/10} -40 = -8-40 = -48 \qquad \text{ m/s} \\\\ a(0) &= \dfrac{ 4}{5 } e^{-0/10} = \dfrac{ 4}{5 } = 0.8 \qquad \text{ m/s}^2 \end{aligned}

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