Answer to Question #229861 in Mechanics | Relativity for jake jornala

Question

. A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward. For this motion, a. What is the distance traveled? b. What is the displacement? c. What is the coach's speed? d. What is the coach's velocity?

Answer 1

Explanations & Calculations

• Jake Jornala this is the very basic of the translational motion in physics so you need to get it clear.
  
• Displacement is the smallest possible straight length measured in between the starting point & the end point of a motion (should be treated directionwise). In contrast the distance travelled is the actual length the object travels no matter the direction. Since both measures a length they both have the same standard unit: meters.

a)

• In this case the total travelled distance is just the sum of those lengths,

$\qquad\qquad \begin{aligned} L&=\small 18m+12m+28m+14m\\ &=\small \bold{72m} \end{aligned}$

b)

• When it comes to the displacement, we need to consider one of the directions to be positive then treat algebraically. Let's take the westward direction to be positive. Then the Eastward direction is set to be negative.
• Ten the displacement or the algebraic sum of the distances is

$\qquad\qquad \begin{aligned} \small d&=\small (+18m)+(-12m)+(+28m)+(-14m)\\ &=\small 18-12+28-14\\ &=\small \bold{+20m} \end{aligned}$

• He is finally located 20m westward from the starting point.

c)

• To assess a speed or velocity a time is needed exactly what we are not given here.

• Suppose we are given a time of $\small t(s)$ for his entire journey, then the speed is given by

$\qquad\qquad \begin{aligned} \small Speed &=\small \frac{\text{Total travelled distance}}{\text{time}}\\ &=\small \frac{72m}{t} \end{aligned}$

d)

• The velocity, again is a vector quantity & is to be treated similar to displacement: directionwise.
• Calculation of the velocity from displacement is done by

$\qquad\qquad \begin{aligned} \small Velocity&=\small \frac{\text{displacement}}{\text{time}}\\ &=\small \frac{20m}{t} \end{aligned}$