Question

two particles start at the origin and move along the x - axis. for 0 ≤ t ≤ 10, their respective position functions are given by (x_1=sin t) and (x_2 = e^{-2t}-1). for how many values of t do the particles have the same velocity?
a none
b one
c two
d three
e four

Explanation:

Step1: Recall velocity - position relationship

Velocity is the derivative of position. So, find the derivatives of $x_1$ and $x_2$.
The derivative of $x_1=\sin t$ with respect to $t$ is $v_1 = \frac{d}{dt}(\sin t)=\cos t$ using the derivative formula $\frac{d}{dt}(\sin t)=\cos t$.
The derivative of $x_2 = e^{-2t}-1$ with respect to $t$ is $v_2=\frac{d}{dt}(e^{-2t}-1)=- 2e^{-2t}$ using the chain - rule $\frac{d}{dt}(e^{u})=e^{u}\cdot\frac{du}{dt}$, where $u = - 2t$ and $\frac{du}{dt}=-2$.

Step2: Set the velocities equal

Set $v_1 = v_2$, so $\cos t=-2e^{-2t}$.
We know that $-1\leqslant\cos t\leqslant1$ and $y = - 2e^{-2t}=-\frac{2}{e^{2t}}$, and the function $y=-2e^{-2t}$ is always negative and its range is $(-2,0)$.
We can use a graphing utility to graph $y = \cos t$ and $y=-2e^{-2t}$ on the interval $0\leqslant t\leqslant10$.
The function $y = \cos t$ oscillates between $- 1$ and $1$ with a period of $2\pi\approx6.28$, and $y=-2e^{-2t}$ is an exponential decay function that starts at $t = 0,y=-2$ and approaches $0$ as $t$ increases.
By graphing $y=\cos t$ and $y = - 2e^{-2t}$ on the interval $[0,10]$, we can see that the two curves intersect three times.

Answer:

D. Three