Question

cfa 4 rev: speed / velocity, dist / displacement and motion graphs
speed

• the equation for speed is _ divided by _.
• for example a car traveling 10 mi in 30 min has a speed of

1st, my equation is $\frac{distance}{time}=\frac{mi}{min}$
but, i have to change min to _, my new equation is $\frac{mi}{hr}$.
my speed is _ mi / hr.
velocity
velocity is _ plus _.
for example in the above speed problem, i can make it velocity by adding a specific direction such as _ _

Explanation:

Step1: Fill in the speed formula

The equation for speed is \( \text{distance} \) divided by \( \text{time} \).

Step2: Calculate speed in the example (first form)

Given distance \(d = 10\) mi and time \(t=30\) min, the initial equation is \(\frac{\text{distance}}{\text{time}}=\frac{10}{30}\) mi/min.

Step3: Convert units

Since \(1\) hour \( = 60\) min, \(30\) min \(=\frac{30}{60}=0.5\) hr. The new equation is \(\frac{10}{0.5}\) mi/hr.

Step4: Calculate the final speed value

\(\frac{10}{0.5}=20\) mi/hr.

Step5: Define velocity

Velocity is \( \text{speed} \) plus \( \text{direction} \). For example, in the above speed problem, it can be made velocity by adding a specific direction such as \( \text{north - east} \) (any specific direction would work).

Answer:

• The equation for speed is \( \text{distance} \) divided by \( \text{time} \).
• The initial equation for the example: \(\frac{10}{30}\) mi/min.
• Convert min to hr: \(30\) min \( = 0.5\) hr, new equation \(\frac{10}{0.5}\) mi/hr.
• Speed is \(20\) mi/hr.
• Velocity is \( \text{speed} \) plus \( \text{direction} \), example direction: \( \text{north - east} \) (or any other specific direction).