# A rocket is accelerating straight up from the surface. At 1.35s after lift, the rocket clears top of its launch platform, 63m above the ground. After additional 4.45s its 1 km above ground. What's the magnitude of average velocity for the 4.45s part?

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what's the magnitude of average velocity for the first 5.80 s of its flight?

what's the magnitude of average velocity for the first 5.80 s of its flight?

##### 1 Answer

Jun 25, 2018

#### Answer:

4.45s part:

5.80s part:

#### Explanation:

Notice that velocity is always in some unit of distance divided by some unit of time.

Average velocity = (total displacement)/(total time)

(Displacement has unit of distance.)

So, to solve the 4.45s part,

To solve the 5.80 s part,

I hope this helps,

Steve