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) = -t^3 +11t^2 -24t \\\\\n\nv(t) = x\u02bc(t) = -3t^2+22t -24"
At t = 7
"v(7) = -3 \\times 7^2 + 22 \\times 7 -24 \\\\\n\n= -147 +154 -24 \\\\\n\n= -17"
Answer: -17
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