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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 ammount 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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