A deer accelerates at a constant rate from 0 mph to 29 mph in 20 seconds. How far does the deer run during this time? A. 0.0806 mi B. 0.1611 mi C. 0.1208 mi D. 0.4833 mi
The correct answer and explanation is:
Correct Answer: B. 0.1611 mi
Step-by-Step Explanation:
We are given:
- Initial velocity vi=0v_i = 0 mph
- Final velocity vf=29v_f = 29 mph
- Time t=20t = 20 seconds
- We are to find the distance in miles that the deer runs during this time while accelerating at a constant rate.
Step 1: Use the average velocity formula
Since the deer is accelerating uniformly, the average speed is: vavg=vi+vf2=0+292=14.5 mphv_{avg} = \frac{v_i + v_f}{2} = \frac{0 + 29}{2} = 14.5 \text{ mph}
Step 2: Convert time from seconds to hours
20 seconds=203600 hours=1180 hours20 \text{ seconds} = \frac{20}{3600} \text{ hours} = \frac{1}{180} \text{ hours}
Step 3: Use the formula for distance
Distance=vavg×t\text{Distance} = v_{avg} \times t Distance=14.5 mph×1180 hr=14.5180≈0.0806 miles\text{Distance} = 14.5 \, \text{mph} \times \frac{1}{180} \, \text{hr} = \frac{14.5}{180} \approx 0.0806 \, \text{miles}
Wait! That gives us option A, but that’s assuming average velocity. Let’s double-check using kinematics in consistent units.
Alternate Method (More Accurate): Use SI Units
Convert 29 mph to meters per second:
29 mph×1609.34 m1 mile×13600 s≈12.96 m/s29 \, \text{mph} \times \frac{1609.34 \, \text{m}}{1 \, \text{mile}} \times \frac{1}{3600 \, \text{s}} \approx 12.96 \, \text{m/s}
Then use the formula: Distance=12vt=12×12.96 m/s×20 s=129.6 m\text{Distance} = \frac{1}{2} v t = \frac{1}{2} \times 12.96 \, \text{m/s} \times 20 \, \text{s} = 129.6 \, \text{m}
Convert meters to miles: 129.6 m×1 mile1609.34 m≈0.0806 mi129.6 \, \text{m} \times \frac{1 \, \text{mile}}{1609.34 \, \text{m}} \approx 0.0806 \, \text{mi}
✅ Confirms that option A is actually correct.
✅ Final Answer: A. 0.0806 mi
The deer runs approximately 0.0806 miles during the 20-second acceleration.