Question

practice exercise 3.2

t₁ = 1.0 s

t₂ = 6.0 s

meters
figure 3.6

instructions: using the motion diagram of a car is shown in figure 3.6. solve for the following:

1. distance
2. displacement
3. speed
4. velocity

Explanation:

1. Distance

Step1: Identify positions

At \( t_1 = 1.0 \, \text{s} \), position \( x_1=- 10 \, \text{m} \); at \( t_2 = 6.0 \, \text{s} \), position \( x_2 = 20 \, \text{m} \).

Step2: Calculate distance

Distance is the total path length, so \( d=\vert x_2 - x_1\vert=\vert20-(- 10)\vert = 30 \, \text{m} \).

2. Displacement

Step1: Recall displacement formula

Displacement \( \Delta x=x_2 - x_1 \).

Step2: Substitute values

\( x_1=-10 \, \text{m} \), \( x_2 = 20 \, \text{m} \), so \( \Delta x=20-(-10)=30 \, \text{m} \).

3. Speed

Step1: Recall speed formula

Speed \( v=\frac{d}{t_2 - t_1} \), where \( t = t_2 - t_1=6.0 - 1.0 = 5.0 \, \text{s} \), \( d = 30 \, \text{m} \).

Step2: Calculate speed

\( v=\frac{30}{5.0}=6.0 \, \text{m/s} \).

4. Velocity

Answer:

s:

1. Distance: \( \boldsymbol{30 \, \text{m}} \)
2. Displacement: \( \boldsymbol{30 \, \text{m}} \) (in the positive direction)
3. Speed: \( \boldsymbol{6.0 \, \text{m/s}} \)
4. Velocity: \( \boldsymbol{6.0 \, \text{m/s}} \) (in the positive direction)