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Answer :
The time interval during which the toy rocket engine provides upward acceleration is approximately 1.60 seconds.
To find the time interval during which the rocket engine provides upward acceleration, we can use the following information: at engine burnout, the rocket has risen to 65.4 m and has a velocity of 50.9 m/s. The relevant kinematic equation for this situation is:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s since the rocket starts from rest), a is the acceleration, and t is the time.
- v = 50.9 m/s
- u = 0 m/s
Using the equation, we can solve for time t:
50.9 m/s = 0 m/s + a(t)
Since the initial velocity is zero, we have:
50.9 m/s = a * t
We still need one more equation that involves the distance covered (which is 65.4 m) during the same time period. We use the equation:
s = ut + 0.5at²
where s is the displacement (65.4 m), u is the initial velocity (0 m/s), a is the acceleration, and t is the time:
65.4 m = 0 * t + 0.5 * a * t²
Simplifying, we get:
65.4 m = 0.5 * a * t²
Therefore:
a * t² = 130.8
Using the relation 50.9 = a * t from earlier, we substitute 'a' to get:
50.9 t = 130.8
t² = 130.8 / 50.9
t² = 2.57
t = √(2.57)
t ≈ 1.60 seconds
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