Question

求解一道微积分应用题。

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

Answer 1

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