(a) To find when the particle's velocity is zero, we need to find the time when the derivative of its position with respect to time is zero. The derivative of x=20t&#x2212;5t3" role="presentation" style="position: relative;" data-evernote-id="187">x=20t−5t3x=20t−5t3 x = 20t - 5t^3  with respect to t" role="presentation" style="position: relative;" data-evernote-id="222">ttt is v=dxdt=20&#x2212;15t2" role="presentation" style="position: relative;" data-evernote-id="236">v=dxdt=20−15t2v=dxdt=20−15t2 v = \frac{dx}{dt} = 20 - 15t^2 . To find when the velocity is zero, we set v" role="presentation" style="position: relative;" data-evernote-id="295">vvv equal to zero and solve for t" role="presentation" style="position: relative;" data-evernote-id="309">ttt:

20&#x2212;15t2=0" role="presentation" style="position: relative;" data-evernote-id="324">20−15t2=020−15t2=0 20 - 15t^2 = 0 

t2=43" role="presentation" style="position: relative;" data-evernote-id="358">t2=43t2=43 t^2 = \frac{4}{3} 

t=&#x00B1;43" role="presentation" style="position: relative;" data-evernote-id="398">t=±43−−√t=±43 t = \pm \sqrt{\frac{4}{3}} 

Therefore, the particle's velocity is zero at t=43&#x2248;1.15" role="presentation" style="position: relative;" data-evernote-id="447">t=43−−√≈1.15t=43≈1.15 t = \sqrt{\frac{4}{3}} \approx 1.15 seconds and t=&#x2212;43&#x2248;&#x2212;1.15" role="presentation" style="position: relative;" data-evernote-id="497">t=−43−−√≈−1.15t=−43≈−1.15t = -\sqrt{\frac{4}{3}} \approx -1.15 seconds.

(b) To find when the particle's acceleration is zero, we need to find the time when the derivative of its velocity with respect to time is zero. The derivative of v" role="presentation" style="position: relative;" data-evernote-id="552">vvv with respect to t" role="presentation" style="position: relative;" data-evernote-id="566">ttt is a=dvdt=&#x2212;30t" role="presentation" style="position: relative;" data-evernote-id="580">a=dvdt=−30ta=dvdt=−30ta = \frac{dv}{dt} = -30t. To find when the acceleration is zero, we set a" role="presentation" style="position: relative;" data-evernote-id="628">aaa equal to zero and solve for t" role="presentation" style="position: relative;" data-evernote-id="642">ttt:

&#x2212;30t=0" role="presentation" style="position: relative;" data-evernote-id="657">−30t=0−30t=0 -30t = 0 

t=0" role="presentation" style="position: relative;" data-evernote-id="680">t=0t=0 t = 0 

Therefore, the particle's acceleration is zero at t=0" role="presentation" style="position: relative;" data-evernote-id="699">t=0t=0t=0 seconds.

(c) The particle's acceleration is negative when a&lt;0" role="presentation" style="position: relative;" data-evernote-id="718">a<0a<0a<0. From part (b), we know that the acceleration is zero at t=0" role="presentation" style="position: relative;" data-evernote-id="736">t=0t=0t=0 seconds. To the left of this time, the particle is still speeding up, so the acceleration is negative. To the right of this time, the particle is slowing down, so the acceleration is also negative. Therefore, the particle's acceleration is negative for all time t" role="presentation" style="position: relative;" data-evernote-id="754">ttt.

(d) The particle's acceleration is positive when a&gt;0" role="presentation" style="position: relative;" data-evernote-id="769">a>0a>0a>0. Since the particle's acceleration is always negative (as we determined in part (c)), it is never positive.
(a) 입자의 속도가 0이 되는 시점을 찾으려면 시간에 대한 입자의 위치 미분이 0이 되는 시간을 구해야 합니다. x=20t&#x2212;5t3" role="presentation" style="위치: 상대적;" data-evernote-id="187">x=20t-5t3x=20t-5t3 x = 20t - 5t^3의 미분은 t에 대해" role="presentation" style="위치": 상대;" data-evernote-id="222">ttt는 v=dxdt=20&#x2212;15t2" role="presentation" style="위치: 상대;" data-evernote-id="236">v=dxdt=20-15t2v=dxdt=20-15t2 v = \frac{dx}{dt} = 20 - 15t^2 . 속도가 0일 때를 구하기 위해 v" role="presentation" style="위치: 상대적;" data-evernote-id="295">vvv를 0으로 설정하고 t" role="presentation" style="위치: 상대적;" data-evernote-id="309">ttt를 풉니다:20&#x2212;15t2=0" role="presentation" style="위치: 상대;" data-evernote-id="324">20-15t2=020-15t2=0 20 - 15t^2 = 0 t2=43" role="presentation" style="위치: 상대적;" data-evernote-id="358">t2=43t2=43 t^2 = \frac{4}{3} t=&#00B1;43" role="presentation" style="위치: 상대적;" data-evernote-id="398">t=±43--√t=±43 t = \pm \sqrt{\frac{4}{3}} 따라서 입자의 속도는 t=43&#x2248;1.15에서 0입니다." role="presentation" style="위치: 상대적;" data-evernote-id="447">t=43--√≈1.15t=43≈1.15 t = \sqrt{\frac{4}{3}}. \약 1.15초 및 t=&#x2212;43&#x2248;&#x2212;1.15" role="presentation" style="위치: 상대;" data-evernote-id="497">t=-43--√≈-1.15t=-43≈-1.15t = -\sqrt{\frac{4}{3}} \약 -1.15초.(b) 입자의 가속도가 0일 때를 구하려면 시간에 대한 속도의 미분이 0이 되는 시간을 구해야 합니다. v의 미분" role="presentation" style="위치: 상대적;" data-evernote-id="552">vvv의 t에 대한 미분" role="presentation" style="위치: 상대;" data-evernote-id="566">ttt는 a=dvdt=&#x2212;30t" role="presentation" style="위치: 상대;" data-evernote-id="580">a=dvdt=-30ta=dvdt=-30ta = \frac{dv}{dt} = -30t입니다. 가속도가 0일 때를 구하기 위해" role="presentation" style="위치: 상대적;" data-evernote-id="628">aaa를 0으로 설정하고 t" role="presentation" style="위치: 상대적;" data-evernote-id="642">ttt를 풀면 됩니다:&#x2212;30t=0" role="presentation" style="위치: 상대적;" data-evernote-id="657">-30t=0-30t=0 -30t = 0 t=0" role="presentation" style="위치: 상대적;" data-evernote-id="680">t=0t=0 t = 0 따라서 입자의 가속도는 t=0에서 0입니다." role="presentation" style="위치: 상대;" data-evernote-id="699">t=0t=0t=0초.(c) 입자의 가속도는 a&lt;0" role="presentation" style="위치: 상대적;" data-evernote-id="718">a<0a<0a<0. 부분 (b)에서 가속도는 t=0" role="presentation" style="위치: 상대적;" data-evernote-id="736">t=0t=0t=0초라는 것을 알 수 있습니다. 이 시간의 왼쪽은 파티클이 여전히 속도를 내고 있으므로 가속도가 음수입니다. 이 시간의 오른쪽은 파티클이 느려지고 있으므로 가속도도 음수입니다. 따라서 파티클의 가속도는 항상 음수입니다. t" role="presentation" style="위치: 상대;" data-evernote-id="754">ttt.(d) 입자의 가속도가 양수일 때 a&gt;0" role="presentation" style="위치: 상대;" data-evernote-id="769">a>0a>0a>0. 입자의 가속도는 항상 음수이므로(부분 (c)에서 결정한 대로) 결코 양이 아닙니다.