Certainly! Let's determine the true speed and direction of the object by combining the given velocity vectors. Here are the steps to solve this problem:
### Step 1: Sum the Components of the Velocity Vectors
Given vectors are:
[tex]\[
v_1 = -60i + 3j
\][/tex]
[tex]\[
v_2 = 4i + 14j
\][/tex]
First, sum the [tex]\(i\)[/tex]-components and [tex]\(j\)[/tex]-components of the two vectors:
[tex]\[
v_{\text{total}, i} = -60 + 4 = -56
\][/tex]
[tex]\[
v_{\text{total}, j} = 3 + 14 = 17
\][/tex]
Thus, the resultant vector is:
[tex]\[
v_{\text{total}} = -56i + 17j
\][/tex]
### Step 2: Calculate the Magnitude (True Speed) of the Resultant Vector
The magnitude or true speed [tex]\(s\)[/tex] of the resultant vector is given by the Euclidean norm:
[tex]\[
s = \sqrt{v_{\text{total}, i}^2 + v_{\text{total}, j}^2}
\][/tex]
Substituting the values:
[tex]\[
s = \sqrt{(-56)^2 + 17^2}
\][/tex]
[tex]\[
s = \sqrt{3136 + 289}
\][/tex]
[tex]\[
s = \sqrt{3425}
\][/tex]
Calculating the square root and rounding to the thousandths place:
[tex]\[
s \approx 58.524
\][/tex]
### Step 3: Calculate the Direction of the Resultant Vector
The direction [tex]\(\theta\)[/tex] of the resultant vector is found using the arctangent function:
[tex]\[
\theta = \arctan\left(\frac{v_{\text{total}, j}}{v_{\text{total}, i}}\right)
\][/tex]
Substituting the values:
[tex]\[
\theta = \arctan\left(\frac{17}{-56}\right)
\][/tex]
Since the [tex]\(i\)[/tex]-component (denominator) is negative and the [tex]\(j\)[/tex]-component (numerator) is positive, the direction is in the second quadrant. We must add 180 degrees to our result to get the correct angle direction in standard position:
[tex]\[
\theta = \arctan\left(\frac{17}{-56}\right) + 180^\circ
\][/tex]
Calculating the arctangent and converting to degrees:
[tex]\[
\theta \approx -17.00^\circ + 180^\circ
\][/tex]
[tex]\[
\theta \approx 163^\circ
\][/tex]
Rounding to the nearest degree, we get 163 degrees.
### Step 4: Conclusion
The true speed of the object is [tex]\(58.524\)[/tex] and the direction is [tex]\(163^\circ\)[/tex]. Thus, the correct answer is:
[tex]\[
\boxed{58.524 ; 163^\circ}
\][/tex]
