scientists released a weather balloon from a raised platform at 4:00 p.m. the weather balloon rose at a…

scientists released a weather balloon from a raised platform at 4:00 p.m. the weather balloon rose at a constant speed. at 4:05 pm, the weather balloons altitude was 1,482 meters. at 4:09, the weather balloon had reached an altitude of 2,626 meters. how many meters did the weather balloon rise each minute? meters how high was the launch platform? meters complete the equation that describes the relationship between the altitude of the weather balloon in meters, a, and the elapsed time in minutes, t. a = t +

scientists released a weather balloon from a raised platform at 4:00 p.m. the weather balloon rose at a constant speed. at 4:05 pm, the weather balloons altitude was 1,482 meters. at 4:09, the weather balloon had reached an altitude of 2,626 meters. how many meters did the weather balloon rise each minute? meters how high was the launch platform? meters complete the equation that describes the relationship between the altitude of the weather balloon in meters, a, and the elapsed time in minutes, t. a = t +

Answer

Explanation:

Step1: Calculate time - interval

From 4:05 pm to 4:09 pm, the time - interval $\Delta t=4:09 - 4:05 = 4$ minutes.

Step2: Calculate altitude - change

The change in altitude $\Delta A=2626 - 1482=1144$ meters.

Step3: Find the speed (meters per minute)

The speed $v=\frac{\Delta A}{\Delta t}=\frac{1144}{4}=286$ meters per minute.

Step4: Find the initial altitude (height of the launch platform)

At 4:05 pm ($t = 5$ minutes after launch), $A = 1482$ meters. Using the equation $A=vt + h_0$ (where $v$ is the speed, $t$ is the time, and $h_0$ is the initial altitude), we substitute $v = 286$, $t = 5$, and $A = 1482$. So, $1482=286\times5+h_0$. Then $1482 = 1430+h_0$, and $h_0=1482 - 1430 = 52$ meters.

Step5: Write the equation

The equation relating altitude $A$ and time $t$ is $A = 286t+52$.

Answer:

286 52 286; 52