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Unformatted text preview: v = -16+9t 2 We set the velocity to be zero and solve the equation for time. -16-9t 2 = 0 t = 1.33 s The position of the particle at t = 1.33 s is x(t = 1.33 s) = 9-16(1.33)+3(1.33)3 = -5.22 m b) What is the particles acceleration when it momentarily stops? The acceleration of the particle is given by the time derivative of the velocity function. We get a = 18t The acceleration at t = 1.33 s is a(t = 1.33 s) = 181.33 = 23.9 m/s 2...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
- Summer '08