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On Moon: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?

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?

Because V(t) = ds/dt = 3t^2 - 12t + 9,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.

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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