A particle moves along the x-axis so that at time t >= 0 (t is greater than or equal to "t") its position is given by x(t)=-t^3+11t^2-24t. Determine the velocity of the particle at t=7.
x(t)=−t3+11t2−24tv(t)=xʼ(t)=−3t2+22t−24x(t) = -t^3 +11t^2 -24t \\ v(t) = xʼ(t) = -3t^2+22t -24x(t)=−t3+11t2−24tv(t)=xʼ(t)=−3t2+22t−24
At t = 7
v(7)=−3×72+22×7−24=−147+154−24=−17v(7) = -3 \times 7^2 + 22 \times 7 -24 \\ = -147 +154 -24 \\ = -17v(7)=−3×72+22×7−24=−147+154−24=−17
Answer: -17
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