Question

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 4 seconds is 59% of its original speed. if during this time it falls a distance of 16,039 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

Explanation:

Step1: Define variables

Let $v_0$ = original speed, $v_f$ = final speed.

Step2: Relate average speed to speeds

Average speed = $\frac{v_0 + v_f}{2} = 0.59v_0$

Step3: Solve for $v_f$

Rearrange: $v_0 + v_f = 1.18v_0$ → $v_f = 1.18v_0 - v_0 = 0.18v_0$

Step4: Find total distance formula

Distance = average speed × time: $16039 = 0.59v_0 \times 4$

Step5: Calculate original speed $v_0$

$0.59v_0 = \frac{16039}{4} = 4009.75$
$v_0 = \frac{4009.75}{0.59} \approx 6796.19$ m/s

Step6: Calculate final speed $v_f$

$v_f = 0.18 \times 6796.19 \approx 1223.31$

Answer:

1223 meters per second