1) v(t) = dx/dt = 3t^2 -42t +144 = 3(t^2 -14t + 48) =3(t-6)(t-8)
a(t) = dv/dt = 6t - 42 =0,t=7
so when t=7,acceleration is zero
at that time,v(t)= 3(7-6)(7-8) = -3 m/s
2) when t=6 or 8,v(t)=0
from t=0 to t=7,the direction of body move changed at t=6
for the first stage,t=0 to t=6,distance = x(6)-x(0) = 6^3 - 21* 6^2 + 144*6
for the second stage,from t=6 to t=7,distance = x(6)-x(7)
you will get the answer if you sum them up
求解一道微積分應用題.
求解一道微積分應用題.
At time t>=0 seconds,the position of a body moving along the x-axis is x(t)=t^3-21t^2+144t m.
(1)Find the body's speed each time the acceleration is zero.Enter your answer as a list if necessary.
(2) Find the total distance traveled by the body from t=0 to t=7
At time t>=0 seconds,the position of a body moving along the x-axis is x(t)=t^3-21t^2+144t m.
(1)Find the body's speed each time the acceleration is zero.Enter your answer as a list if necessary.
(2) Find the total distance traveled by the body from t=0 to t=7
英語人氣：674 ℃時間：2020-06-20 01:20:24
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