# 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?

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

Jun 25, 2018

4.45s part: ${v}_{\text{ave}} = 210.6 \frac{m}{s}$

5.80s part: ${v}_{\text{ave}} = 172.4 \frac{m}{s}$

#### 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,

${v}_{\text{ave}} = \frac{1000 m - 63 m}{4.45 s} = 210.6 \frac{m}{s}$

To solve the 5.80 s part,

${v}_{\text{ave}} = \frac{1000 m}{5.80 s} = 172.4 \frac{m}{s}$

I hope this helps,
Steve