Question

suppose the velocity, v, of an object is given by $v = 2t^{2}+13t - 15$, where t is time in seconds. (here t can be positive or negative.) find the intervals where the velocity is negative. choose the correct solution below. a. less than or equal to 0.1 sec and more than or equal to 2 sec b. between - 7.5 sec and 1 sec c. less than 0.5 sec and more than 15 sec d. between - 2 sec and 15 sec

Explanation:

Step1: Solve the quadratic inequality \(2t^{2}+13t - 15<0\)

Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 2\), \(b=13\), \(c=-15\).

$$t=\frac{-13\pm\sqrt{13^{2}-4\times2\times(-15)}}{2\times2}=\frac{-13\pm\sqrt{169 + 120}}{4}=\frac{-13\pm\sqrt{289}}{4}=\frac{-13\pm17}{4}$$

We get \(t_1=\frac{-13 + 17}{4}=1\) and \(t_2=\frac{-13-17}{4}=-7.5\)

Step2: Determine the intervals

The quadratic function \(y = 2t^{2}+13t - 15\) is a parabola opening upwards (\(a=2>0\)). The inequality \(2t^{2}+13t - 15<0\) holds for values of \(t\) between the two roots.

Answer:

B. between \(-7.5\) sec and \(1\) sec