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qz1sol_3891f10

# qz1sol_3891f10 - v =-16 9t 2 We set the velocity to be zero...

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TA: Tomoyuki Nakayama Tuesday, August 31, 2010 PHY 2048: Physic 1, Discussion Section 3891 Quiz 1 (Homework Set #1) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A particle position is given by x = 9–16 t +3 t 3 , in which x is in meters and t is in seconds. a) Where is the particle when it momentarily stops? The velocity of the particle is given by the time derivative of the position function. We have
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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 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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