a meteor enters the earths atmosphere and slows due to air drag at a constant rate such that its average…

a meteor enters the earths atmosphere and slows due to air drag at a constant rate such that its average speed for a period of 3 seconds is 65% of its original speed. if during this time it falls a distance of 16,278 meters, find the speed of the meteor in the atmosphere, at the end of this time period, in meters per second. you may round to the nearest whole number
Answer
Explanation:
Step1: Find the average speed
Use the formula (v_{avg}=\frac{d}{t}). Given (d = 16278) meters and (t=3) seconds. (v_{avg}=\frac{16278}{3}=5426) m/s.
Step2: Relate average speed to original speed
Let the original speed be (v_0). Given (v_{avg}=0.65v_0). Since (v_{avg} = 5426) m/s, then (v_0=\frac{5426}{0.65}=8347.6923) m/s.
Step3: Use the formula for average speed in uniformly - decelerated motion
The formula for average speed in uniformly - decelerated motion is (v_{avg}=\frac{v_0 + v}{2}), where (v) is the final speed. We know (v_{avg}=5426) m/s and (v_0 = 8347.6923) m/s. Substitute into the formula: (5426=\frac{8347.6923 + v}{2}). Multiply both sides by 2: (10852=8347.6923 + v). Solve for (v): (v=10852 - 8347.6923=2504.3077\approx2504) m/s.
Answer:
(2504)