Chapter 10, Problem 52EP

a)

To determine

To find: The velocity vector for each particle at time $t=3$ .

Answer to Problem 52EP

At $t=3$ , the velocity vector for particle A is $〈1,2〉$ and for particle B is $〈\frac{3}{2},\frac{3}{2}〉$ .

Explanation of Solution

Given information:

The position of particle A is given by the equations $x=t-2$ and $y={\left(t-2\right)}^2$ .

The position of particle B is given by the equations $x=\frac{3}{2}t-4$ and $y=\frac{3}{2}t-2$ .

Formulas used:

Suppose a particle moves along a smooth curve in the plane and its position at time $t$ is $\left(x\left(t\right),y\left(t\right)\right)$ . Then the position vector of the particle is $r\left(t\right)=〈x\left(t\right),y\left(t\right)〉$ and the velocity vector of the particle is $v\left(t\right)=〈\frac{dx}{dt},\frac{dy}{dt}〉$ .

Calculation:

The velocity vector of the particle is $v\left(t\right)=〈\frac{dx}{dt},\frac{dy}{dt}〉$ at $t=3$ .

Differentiate equation $x=t-2$ for $t$ .

  $\frac{dx}{dt}=1$

Differentiate equation $y={\left(t-2\right)}^2$ for $t$ .

  $\frac{dy}{dt}=2\left(t-2\right)$

Substitute $3$ for $t$ the equation $〈\frac{dx}{dt},\frac{dy}{dt}〉=〈1,2\left(t-2\right)〉$ to find velocity of particle A.

  $\begin{array}{c}〈\frac{dx}{dt},\frac{dy}{dt}〉=〈1,2\left(\left(3\right)-2\right)〉\\ =〈1,2〉\end{array}$

Differentiate equation $x=\frac{3}{2}t-4$ for $t$ .

  $\frac{dx}{dt}=\frac{3}{2}$

Differentiate equation $y=\frac{3}{2}t-2$ for $t$ .

  $\frac{dy}{dt}=\frac{3}{2}$

The velocity of particle B at $t=3$ is $〈\frac{dx}{dt},\frac{dy}{dt}〉=〈\frac{3}{2},\frac{3}{2}〉$ .

Thus, the velocity vector of particle A at $t=3$ is $〈1,2〉$ and particle B at $t=3$ is $〈\frac{3}{2},\frac{3}{2}〉$ .

b)

To determine

To find: The distance traveled by particle A from $t=0$ to $t=3$ .

Answer to Problem 52EP

The distance traveled by particle A from $t=0$ to $t=3$ is $6.126$ .

Explanation of Solution

Given information: The position of particle A is given by $x=t-2$ and $y={\left(t-2\right)}^2$ .

Formulas used:

Suppose a particle moves along a smooth curve in the plane and its position at time $t$ is $\left(x\left(t\right),y\left(t\right)\right)$ . Then the velocity vector of the particle is $v\left(t\right)=〈\frac{dx}{dt},\frac{dy}{dt}〉$ and the speed of the particle is $\left|v\left(t\right)\right|$ .

The distance traveled by a particle from $t=a$ to $t=b$ is given by the formula ${\displaystyle {\int }_a^b\left|v\left(t\right)\right|dt}={\displaystyle {\int }_a^b\sqrt{{\left(x^{\prime }\left(t\right)\right)}^2+{\left(y^{\prime }\left(t\right)\right)}^2}dt}$ .

Calculation:

Differentiate equation $x=t-2$ for $t$ .

  $\frac{dx}{dt}=1$

Differentiate equation $y={\left(t-2\right)}^2$ for $t$ .

  $\frac{dy}{dt}=2\left(t-2\right)$

Find the distance traveled by particle A from $t=0$ to $t=3$ .

  $\begin{array}{c}{\displaystyle {\int }_a^b\left|v\left(t\right)\right|dt}={\displaystyle {\int }_0^3\sqrt{{\left(1\right)}^2+{\left(2\left(t-2\right)\right)}^2}dt}\\ ={\displaystyle {\int }_0^3\sqrt{1+{\left(2t-4\right)}^2}dt}\\ ={\left.\frac{1}{2}\left[\frac{1}{2}x\sqrt{x^2+1}+\frac{1}{2}\log \left|x+\sqrt{x^2+1}\right|\right]\right|}_{-4}^2\left(\text{Take }x=2t-4,x=-4\text{ to }2\right)\\ =6.126\end{array}$

Thus, the distance traveled by particle A from $t=0$ to $t=3$ is $6.126$ .

c)

To determine

To find: The exact time when the particle collides.

Answer to Problem 52EP

Both the particles collide at $t=4$ .

Explanation of Solution

Given information: The position of particle A is given by $x=t-2$ and $y={\left(t-2\right)}^2$ .

The position of particle B is given by $x=\frac{3}{2}t-4$ and $y=\frac{3}{2}t-2$ .

Calculation:

If the particles collide then the position vector is the same for both the particles.

Thus, $x_A\left(t\right)=x_B\left(t\right)$ and $y_A\left(t\right)=y_B\left(t\right)$ .

If $x_A\left(t\right)=x_B\left(t\right)$:

  $\begin{array}{c}{x}_A\left(t\right)=x_B\left(t\right)\\ t-2=\frac{3}{2}t-4\\ -\frac{1}{2}t=-2\\ t=4\end{array}$

Substitute 4 for $t$ in $y_A\left(t\right)={\left(t-2\right)}^2$ .

  $\begin{array}{c}{y}_A\left(t\right)={\left(4-2\right)}^2\\ =4\end{array}$

Substitute 4 for $t$ in $y_B\left(t\right)=\frac{3}{2}t-2$ .

  $\begin{array}{c}{y}_B\left(t\right)=\frac{3}{2}\left(4\right)-2\\ =4\end{array}$

At $t=4$ , $y_A\left(t\right)=y_B\left(t\right)$ .

Thus, both the particles collide at $t=4$ .