Unformatted text preview: v = 16+9t 2 We set the velocity to be zero and solve the equation for time. 169t 2 = 0 ⇒ t = 1.33 s The position of the particle at t = 1.33 s is x(t = 1.33 s) = 916(1.33)+3(1.33)3 = 5.22 m b) What is the particle’s 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) = 18×1.33 = 23.9 m/s 2...
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 Summer '08
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 Derivative, Work, Velocity, Particle, Expression, Tomoyuki Nakayama

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