Question

A rocket with a satellite accelerates upward from the earth's surface. At 1.15 sec. after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. What is the magnitude of the average velocity of the rocket for the 4.75 s flight?

Answer 1

The average velocity of the rocket during the first 4.75s is 226.75m/s upwards.

Explanation:

Note: I'm assuming a constant acceleration for this problem.

We can solve this problem using basic kinematics.

Since the rocket was at rest when launched, our initial velocity is `0m/s` and we'll define the upwards direction as positive.

Find the acceleration of the rocket using our first set of info
`d=v_it+1/2at^2`
`a=2(d-v_it)/t^2 = 2(63m-0)/(1.15s)^2 ~~ 95.27m/s^2`

Find the final velocity of the rocket at 4.75s
`v_f=v_i+at = 0+(95.27m/s^2)(4.75s) ~~ 452.55m/s`

Find the average velocity of the rocket for the first 4.75s
`bar v = (v_i+v_f)/2 = (0+452.55m/s)/2 ~~ 226.75m/s`

Since we're looking for an average velocity and not an average speed we also need to include a direction, so our final answer is `226.75m/s` upwards.

Answer 2

`bar(v) = "197 m/s"`

Explanation:

FULL QUESTION

A rocket with a satellite accelerates upward from the Earth's surface. At 1.15 sec. after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 sec, it is 1.00 km above the ground.

What is the magnitude of the average velocity of the rocket for the 4.75 s flight?

So, you know that your rocket covers 63 m in 1.15 s, which means that you can determine the speed with which it clears the platform by using

`h = v_0 * t + 1/2 * a * t^2`

Since the rocket started at rest, its initial velocity will be equal to zero. The acceleration of the rocket for the first 1.15 s of flight will be

`h = 1/2 * a * t^2`

`a = (2h)/t^2 = (2 * "63 m")/(1.15""^2 "s"^2) = "95.3 m/s"""^2`

The rocket's velocity after 1.15 s of flight is

`v_1 = underbrace(v_0)_(color(blue)("=0")) * t + a * t`

`v_1 = 95.3"m"/"s"^color(red)(cancel(color(black)("2"))) * 1.15color(red)(cancel(color(black)("s"))) = "109.6 m/s"`

Now, you know that after an additional 4.75 seconds, the rocket is at a height of 1.00 km above the ground.

This means that it covered, in 4.75 seconds of flight, an additional

`h_2 = 1000 - 63 = "937 m"`

Use this distance and the flight time to determine its acceleration for the second portion of the flight

`h_2 = underbrace(v_0)_(color(blue)(=v_1)) * t_2 + 1/2 * a_2 * t_2^2`

`a_2 = (2 * ( h_2 - v_1 * t_2))/t_2^2`

`a_2 = (2 * ("937 m" - 109.6"m"/color(red)(cancel(color(black)("s"))) * 4.75color(red)(cancel(color(black)("s")))))/(4.75^2"s"^2)`

`a_2 = (2 * "416.4 m")/("22.6 s"^2) = "36.8 m/s"""^2`

This means tha the speed of the rocket after the second part of its flight will be

`v_2 = v_1 + a_2 * t_2`

`v_2 = 109.6"m"/"s" + 36.8"m"/color(red)(cancel(color(black)("s"))) * 4.75color(red)(cancel(color(black)("s"))) ="284.4 m/s"`

Therefore, the magnitude of the average velocity is

`bar(v) = (v_1 + v_2)/2 = (109.6 + 284.4)/2 = color(green)("197 m/s")`