Question 942 of 949
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of 200 meters. What is the car's acceleration?
- 1.56 m/s²
- 3.12 m/s²
- 6.25 m/s²
- 12.5 m/s²
Correct Answer:
B
Explanation
To find the car's acceleration, we can use one of the equations of motion that relates initial velocity, final velocity, acceleration, and distance. The equation we will use is:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (25 m/s)
- \( u \) = initial velocity (0 m/s, since the car starts from rest)
- \( a \) = acceleration (what we are trying to find)
- \( s \) = distance (200 m)
### Step-by-Step Solution:
1. **Identify the known values**:
- Final velocity, \( v = 25 \, \text{m/s} \)
- Initial velocity, \( u = 0 \, \text{m/s} \)
- Distance, \( s = 200 \, \text{m} \)
2. **Substitute the known values into the equation**:
\[
(25 \, \text{m/s})^2 = (0 \, \text{m/s})^2 + 2a(200 \, \text{m})
\]
3. **Calculate \( v^2 \)**:
\[
625 \, \text{m}^2/\text{s}^2 = 0 + 400a
\]
4. **Rearrange the equation to solve for \( a \)**:
\[
625 = 400a
\]
\[
a = \frac{625}{400}
\]
5. **Perform the division**:
\[
a = 1.5625 \, \text{m/s}^2
\]
6. **Round to two decimal places**:
\[
a \approx 1.56 \, \text{m/s}^2
\]
### Conclusion:
The correct answer is **A. 1.56 m/s²**.
### Explanation of Other Options:
- **Option B (3.12 m/s²)**: This value is incorrect because it likely results from a miscalculation or misunderstanding of the relationship between distance, velocity, and acceleration. It does not satisfy the equation of motion used.
- **Option C (6.25 m/s²)**: This value is also incorrect. It may arise from incorrectly assuming a different relationship or misapplying the formula. It does not fit the context of the problem.
- **Option D (12.5 m/s²)**: This option is significantly higher than the calculated acceleration. It could be a result of misunderstanding the units or the relationship between the variables involved.
### Common Pitfalls:
- **Forgetting to square the final velocity**: When using the equation \( v^2 = u^2 + 2as \), it’s crucial to remember that \( v \) must be squared.
- **Misunderstanding the initial conditions**: Always ensure that the initial velocity is correctly identified, especially if the object starts from rest.
- **Not rearranging the equation correctly**: When solving for acceleration, ensure that all terms are correctly isolated.
### Revision Summary:
- Use the equation \( v^2 = u^2 + 2as \) to relate velocity, acceleration, and distance.
- Substitute known values carefully and perform calculations step-by-step.
- Be cautious of common mistakes, such as squaring values and rearranging equations.
- The final answer for the car's acceleration in this scenario is **1.56 m/s²**.