Answer:
equilibrium position.
Explanation:
In simple harmonic motion , velocity v(t) is given by,
v(t) = -ω A sin(ωt + φ)
where
ω = angular velocity of the corresponding circular motion
A = amplitude
t = time
φ = the initial angle of the corresponding circular motion when the motion begin.
v (t) get maximized when sin value is maximized , i.e. sin =1
The particle has maximum speed when it passes through the equilibrium position.
Answer:
10s
Explanation:
Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.
If acceleration is constant, then the velocity is changing by a constant amount.
With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:
where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.
So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.
<u>Verification:</u>
In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.
Recognizing that the acceleration is , the velocity increases by 100 units [m/s] every second.
At time 0[s], the velocity is 0[m/s]
At time 1[s], the velocity is 100[m/s]
At time 2[s], the velocity is 200[m/s]
At time 3[s], the velocity is 300[m/s]
At time 4[s], the velocity is 400[m/s]
At time 5[s], the velocity is 500[m/s]
At time 6[s], the velocity is 600[m/s]
At time 7[s], the velocity is 700[m/s]
At time 8[s], the velocity is 800[m/s]
At time 9[s], the velocity is 900[m/s]
At time 10[s], the velocity is 1000[m/s]
So, indeed, after 10 seconds, the velocity reaches 1000 m/s