A particle P moves in a straight line such that t seconds after passing a fixed point Q, its velocity i given by the equation
V=2t^2-10t+12 . Find:
(a). The values of t when P is instantly at rest
(b) An expression for the distance moved by P after t seconds.
(c). The total distance traveled by P in the first 3 seconds after passing point Q
(d). The distance of P from Q when acceleration is zero
(a) When P is instantly at rest, its velocity is zero. Therefore, we need to solve for t in the equation: 2t^2 - 10t + 12 = 0. This is a quadratic equation, and we can factor it as: 2(t-1)(t-6) = 0. Therefore, the values of t when P is instantly at rest are t = 1 and t = 6.
(b) To find the distance moved by P after t seconds, we need to integrate its velocity function from 0 to t: d(t) = ∫ V dt (from 0 to t). Integrating V with respect to t, we get: d(t) = (2/3)t^3 - (5/2)t^2 + 12t + C, where C is a constant of integration. To determine C, we can use the initial condition d(0) = 0, since P starts at Q: 0 = C, so the expression for the distance moved by P is: d(t) = (2/3)t^3 - (5/2)t^2 + 12t.
(c) The total distance traveled by P in the first 3 seconds after passing Q is the absolute value of the displacement, which is the difference between its location at time t = 0 and t = 3: |d(3) - d(0)|. Using the expression for d(t) from part (b), we have: |d(3) - d(0)| = |(2/3)(3)^3 - (5/2)(3)^2 + 12(3) - 0| = 15.
(d) When acceleration is zero, the velocity function V(t) reaches a maximum or minimum value. We can find these values by taking the derivative of V with respect to t and setting it equal to zero: V'(t) = 4t - 10 = 0, which gives t = 2.5. Therefore, P is accelerating towards Q for t < 2.5, and accelerating away from Q for t > 2.5. To find the distance of P from Q when acceleration is zero, we need to find d(2.5), which is: d(2.5) = (2/3)(2.5)^3 - (5/2)(2.5)^2 + 12(2.5) = 10.75. Therefore, the distance of P from Q when acceleration is zero is 10.75 units.