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?

Question

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

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Answer 1 · 2018-06-25T23:53:45

4.45s part: $v_{ave} = 210.6 \frac{m}{s}$ $v_"ave" = 210.6 m/s$

5.80s part: $v_{ave} = 172.4 \frac{m}{s}$ $v_"ave" = 172.4 m/s$

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,

$v_{ave} = \frac{1000 m - 63 m}{4.45 s} = 210.6 \frac{m}{s}$ $v_"ave" = (1000 m - 63m)/(4.45 s) = 210.6 m/s$

To solve the 5.80 s part,

$v_{ave} = \frac{1000 m}{5.80 s} = 172.4 \frac{m}{s}$ $v_"ave" = (1000 m)/(5.80 s) = 172.4 m/s$

I hope this helps,
Steve