+86 A particle moves so that it is at \((2t + 7, 4t - 13)\) at time \(t.\) Find the speed of the particle, measured in unit of distance per unit of time.
+570 The position of the particle is given by the vector (2t+7,4−13t). Taking the derivative of this vector with respect to time t gives us the velocity vector (2,−13).
Speed is the magnitude of the velocity vector, which is found by taking the square root of the sum of the squares of the entries in the velocity vector.
Therefore, the speed is [||(2, -13)|| = \sqrt{2^2 + (-13)^2} = \sqrt{169 + 4} = \sqrt{173}.]
However, speed is a scalar quantity and so it does not have a direction. Therefore, the speed of the particle is sqrt(173) units per unit of time.
