Question

How do I calculate the acceleration, distance traveled and average velocity of an object that starts at 38.9 m/s and stops in 4.4 seconds?

Answer 1

To calculate acceleration, distance traveled, and average velocity for an object that starts at a certain speed and comes to a stop, we can use the following formulas from kinematics.

### Given:
- Initial velocity (\( v_0 \)): 38.9 m/s
- Final velocity (\( v_f \)): 0 m/s (the object stops)
- Time (\( t \)): 4.4 seconds

### 1. Calculate Acceleration

Acceleration (\( a \)) can be calculated using the formula:
\[
a = \frac{v_f - v_0}{t}
\]

Plugging in the values:
\[
a = \frac{0 \, \text{m/s} - 38.9 \, \text{m/s}}{4.4 \, \text{s}} = \frac{-38.9 \, \text{m/s}}{4.4 \, \text{s}} \approx -8.84 \, \text{m/s}^2
\]
The negative sign indicates that the object is decelerating.

### 2. Calculate Distance Traveled

The distance traveled (\( d \)) can be calculated using the formula:
\[
d = v_0 t + \frac{1}{2} a t^2
\]

Substituting the known values:
\[
d = (38.9 \, \text{m/s})(4.4 \, \text{s}) + \frac{1}{2}(-8.84 \, \text{m/s}^2)(4.4 \, \text{s})^2
\]
Calculating each part:
1. \( 38.9 \, \text{m/s} \times 4.4 \, \text{s} = 171.16 \, \text{m} \)
2. \( \frac{1}{2}(-8.84 \, \text{m/s}^2)(4.4^2) \)
- Calculate \( 4.4^2 = 19.36 \, \text{s}^2 \)
- Now calculate \( \frac{1}{2}(-8.84)(19.36) = -85.51 \, \text{m} \)

Now combine these:
\[
d = 171.16 \, \text{m} - 85.51 \, \text{m} = 85.65 \, \text{m}
\]

### 3. Calculate Average Velocity

Average velocity (\( v_{avg} \)) can be calculated using:
\[
v_{avg} = \frac{v_0 + v_f}{2}
\]
Substituting the values:
\[
v_{avg} = \frac{38.9 \, \text{m/s} + 0 \, \text{m/s}}{2} = \frac{38.9 \, \text{m/s}}{2} = 19.45 \, \text{m/s}
\]

### Summary of Results
- Acceleration: \( \approx -8.84 \, \text{m/s}^2 \)
- Distance Traveled: \( \approx 85.65 \, \text{m} \)
- Average Velocity: \( 19.45 \, \text{m/s} \)