Chapter 11.2, Problem 58E

To determine

Todetermine:Thecorrect option for the speed of the particle when $t=2$ from (A) 1.130 (B) 3.018 (C) $\langle -1.307,2.720\rangle$ (D) $\langle 0.757,0.839\rangle$ (E) $\langle 1.307,2.720\rangle$ .

Answer to Problem 58E

The correct option is (B).

Explanation of Solution

Given information:

  $x=\sin 2t$

  $y=\cos 5t$

  $t=2$

Formula Used:

Velocity vector of an object with position vector $\langle x\left(t\right),y\left(t\right)\rangle$ at time $t$ is:

  $v\left(t\right)=\langle \frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right)\rangle$

Pythagoras theorem for a right angle triangle:

  $c=\sqrt{a^2+b^2}$ , $c=\text{hypotenuse},a=\text{adjecent}\text{side},b=\text{opposite}\text{side}$ .

Calculation:

Let the position vector is $r\left(t\right)$ .

Thus $r\left(t\right)=\langle x,y\rangle$

Now, calculate the velocity vector. Use the formula $v\left(t\right)=\langle \frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right)\rangle$ .

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}r\left(t\right)\\ =\frac{d}{dt}\langle sin2t,\cos 5t\rangle \\ =\langle \frac{d}{dt}sin2t,\frac{d}{dt}\cos 5t\rangle \\ =\langle 2cos2t,-5sin5t\rangle \end{array}$

Substitute the value 2 for $t$ .

  $\begin{array}{c}v\left(2\right)=\langle 2cos2t,-5sin2t\rangle \\ =\langle 2cos4,-5sin10\rangle \\ =\langle -1.307,2.720\rangle \end{array}$

Since, speed is a scaler quantity. Thus it has only the magnitude not the direction.

Now, calculate speed. Use Pythagoras theorem $c=\sqrt{a^2+b^2}$

  $\begin{array}{c}\text{speed}=\sqrt{{\left(-1.307\right)}^2+{\left(2.720\right)}^2}\\ =3.018\end{array}$

Conclusion:

Hence, the correct option is (B).