Question
11 A bullet travelling at \( 2 \times 10^{3} \mathrm{~km} \mathrm{~h}^{-1} \) hits a target 500 m away. Find the time of the bullet's flight, in seconds.
Ask by Edwards Warner. in Australia
Feb 04,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The bullet's flight time is approximately 0.9 seconds.
Solution
To find the time of the bullet's flight, we can use the formula for time, which is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
Given:
- Distance to the target = 500 m
- Speed of the bullet = \( 2 \times 10^{3} \mathrm{~km} \mathrm{~h}^{-1} \)
First, we need to convert the speed from kilometers per hour to meters per second.
1 kilometer = 1000 meters
1 hour = 3600 seconds
Therefore, the speed in meters per second is:
\[ 2 \times 10^{3} \mathrm{~km} \mathrm{~h}^{-1} = 2 \times 10^{3} \times \frac{1000}{3600} \mathrm{~m} \mathrm{~s}^{-1} \]
Now, we can calculate the time of the bullet's flight using the formula.
Calculate the value by following steps:
- step0: Calculate:
\(\frac{500}{\left(\frac{2\times 10^{3}\times 1000}{3600}\right)}\)
- step1: Remove the parentheses:
\(\frac{500}{\frac{2\times 10^{3}\times 1000}{3600}}\)
- step2: Reduce the fraction:
\(\frac{500}{\frac{5000}{9}}\)
- step3: Multiply by the reciprocal:
\(500\times \frac{9}{5000}\)
- step4: Reduce the numbers:
\(1\times \frac{9}{10}\)
- step5: Multiply:
\(\frac{9}{10}\)
The time of the bullet's flight is approximately 0.9 seconds.
Reviewed and approved by the UpStudy tutoring team
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Beyond the Answer
To find the time of the bullet's flight, we first need to convert the bullet's speed from kilometers per hour to meters per second. Given the speed is \( 2 \times 10^{3} \text{ km/h} \): \[ 2 \times 10^{3} \text{ km/h} = \frac{2 \times 10^{3} \times 1000 \text{ m}}{3600 \text{ s}} \approx 555.56 \text{ m/s} \] Now, using the formula for time, we have: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{500 \text{ m}}{555.56 \text{ m/s}} \approx 0.9 \text{ seconds} \] So, the time of the bullet's flight is approximately \(0.9\) seconds.