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PrudensOptimus
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A bullet fired straight up from the moon's surface would reach a height of s = 832t - 2.6t^2 after t sec. On Earth, in the absence of air, its height would be s = 832t - 16t^2 after t sec. How long would it take the bullete to get back down in each case?
On Moon:
It would take the same amount of time to get down as it took to go up(maximum displacement, when v = 0).
Thus, V(t) = ds/dt = 832 - 5.2t = 0. t = 160s. It would take about 2 minutes and 40 seconds.
On Earth:
Same notion. V(t) = ds/dt = 832 - 32t = 0. t = 26s.
-- Am I right?
Question 2:
The position of a body at time t sec is s = t^3 - 6t^2 +9t meters. Find the body's acceleration each time the velocity is 0.
Because V(t) = ds/dt = 3t^2 - 12t + 9,
a(t) = dv/dt = 6t - 12
Particle has v = 0 at t = 3, and 1 sec
Thus, the particle has acelleration at v = 0 at:
a(3) = 6 m/s^2
and
a(1) = -6 m/s^2
-- Am I right?
Please correct my mistake. Thanks.
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