During the second stage of a rocket launch the rockets upward velocity increases from 980 m/s to 9000 m/s with an average acceleration of 33 m/s*.
How long did the acceleration last??
I'm stuck on this same question. What I can't figure out is how to solve for t?
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a=33m/s*2 Vf=9000m/s Vi=980m/s ^T=?
^T=Vf-Vi/a
^T=9000m/s-980m/s/33m/s*2 = 8020m/s/33m/s*2
^T=243seounds.= 243/60seounds=4minutes.
The acceleration lasted for 4minutes.
To find the time duration of the acceleration, you can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = average acceleration
t = time duration
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Given:
u = 980 m/s
v = 9000 m/s
a = 33 m/s²
Substituting these values into the equation:
t = (9000 - 980) / 33
Calculating this expression gives us:
t = 8010 / 33
t ≈ 242.73 seconds
Therefore, the acceleration lasted approximately 242.73 seconds.
Use
v(t) = v(0) + ā t
Where
v(0)=980[m/s]
v(t)=9000[m/s]
ā=33[m/s²]
So:
9000 = 980 + 33 t
Solve for t