e) A body moves in a straight line so that its distance S meters from a fixed point 0 at time t seconds is given by S=(t-2)^2(2t-7) Find the time when the body passes through point 0 and the acceleration each time it passes through 0. (3 Marks)

To find the time when the body passes through point 0, we need to set the distance S equal to 0 and solve for t.<br /><br />$S = (t-2)^2(2t-7) = 0$<br /><br />This equation is satisfied when either $(t-2)^2 = 0$ or $(2t-7) = 0$.<br /><br />$(t-2)^2 = 0$ gives $t = 2$.<br /><br />$(2t-7) = 0$ gives $t = \frac{7}{2}$.<br /><br />So, the body passes through point 0 at $t = 2$ seconds and $t = \frac{7}{2}$ seconds.<br /><br />To find the acceleration each time the body passes through 0, we need to take the second derivative of the position function S(t) with respect to t.<br /><br />$S(t) = (t-2)^2(2t-7)$<br /><br />Taking the first derivative:<br /><br />$S'(t) = 2(t-2)(2t-7) + (t-2)^2(2)$<br /><br />$S'(t) = 2(t-2)(2t-7) + 2(t-2)^2$<br /><br />$S'(t) = 2(t-2)(2t-7 + (t-2))$<br /><br />$S'(t) = 2(t-2)(3t-9)$<br /><br />Taking the second derivative:<br /><br />$S''(t) = 2(3t-9) + 2(t-2)(3)$<br /><br />$S''(t) = 6t-18 + 6(t-2)$<br /><br />$S''(t) = 6t-18 + 6t-12$<br /><br />$S''(t) = 12t-30$<br /><br />Now, we can evaluate the acceleration at the times when the body passes through point 0:<br /><br />At $t = 2$ seconds:<br /><br />$S''(2) = 12(2)-30 = 24-30 = -6$ m/s²<br /><br />At $t = \frac{7}{2}$ seconds:<br /><br />$S''(\frac{7}{2}) = 12(\frac{7}{2})-30 = 42-30 = 12$ m/s²<br /><br />Therefore, the acceleration each time the body passes through point 0 is -6 m/s² at $t = 2$ seconds and 12 m/s² at $t = \frac{7}{2}$ seconds.