(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.
(b) The velocity and acceleration at any time t is v = (4ti + j) m/s and a = a = 4i m/s²
(c) The average acceleration in the time interval given in part (a) is 3.98 m/s².
<h3>Position of the particle</h3>
x = at²i + btj
x = 2t²i + tj
<h3>Average velocity, at t₁=2sec and t₂=3sec</h3>
Δv = Δx/Δt
x(2) = 2(2)²i + 2j
x(2) = 8i + 2j
|x(2)| = √(8² + 2²) = 8.246
x(3) = 2(3)²i + 3j
x(3) = 18i + 3j
|x(3)| = √(18² + 3²) = 18.248
Δv = (18.248 - 8.246)/(3 - 2)
Δv = 10 m/s
<h3>Velocity and acceleration at any time, t</h3>
v = dx/dt
v = (4ti + j) m/s
a = dv/dt
a = 4i m/s²
<h3>Average acceleration</h3>
v(2) = 4(2)i + j
v(2) = 8i + j
|v(2)| = 8.06 m/s
v(3) = 4(3)i + j
v(3) = 12i + j
|v(3)| = 12.04 m/s
a = (12.04 - 8.06)/(3 - 2)
a = 3.98 m/s²
Learn more about average acceleration here: brainly.com/question/104491
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Answer:
330Ns
Explanation:
Given parameters:
Mass of the box = 5kg
Initial velocity = 0m/s
Force applied = 15N
Time = 22s
Unknown:
Impulse = ?
Final momentum = ?
Solution:
The impulse of the box can be determined using the expression below:
Impulse = Force x time
Impulse = 15 x 22 = 330Ns
The final momentum is the same as this impulse
The impulse and the final momentum is 330Ns
I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors. Yes, they land in the ocean and the US Navy retrieves them for later use.
