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a particle has a constant acceleration of $6.2\\ \text{m/s}^2$. (due to the nature of this problem, do not use rounded intermediate values—including answers submitted in webassign—in your calculations.)
(a) if its initial velocity is $2.5\\ \text{m/s}$, at what time (in s after $t = 0$) is its displacement $3.2\\ \text{m}$?
(b) what is its speed at that time (in m/s)?

Explanation:

Step1: List known values, use kinematic eq.

Known: $u=2.5\ \text{m/s}$, $a=6.2\ \text{m/s}^2$, $s=3.2\ \text{m}$. Use $s=ut+\frac{1}{2}at^2$.
Substitute values:
$$3.2 = 2.5t + \frac{1}{2}(6.2)t^2$$
Simplify:
$$3.1t^2 + 2.5t - 3.2 = 0$$

Step2: Solve quadratic equation $ax^2+bx+c=0$

Quadratic formula: $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=3.1$, $b=2.5$, $c=-3.2$
Calculate discriminant:
$$\Delta = 2.5^2 - 4(3.1)(-3.2) = 6.25 + 39.68 = 45.93$$
$$t=\frac{-2.5+\sqrt{45.93}}{2(3.1)}$$
(We take the positive root, as time can't be negative)
$$\sqrt{45.93}\approx6.777$$
$$t=\frac{-2.5+6.777}{6.2}=\frac{4.277}{6.2}\approx0.690\ \text{s}$$

Step3: Find speed using $v=u+at$

Substitute $u=2.5$, $a=6.2$, $t\approx0.690$:
$$v=2.5 + (6.2)(0.690)$$
$$v=2.5 + 4.278=6.778\approx6.78\ \text{m/s}$$

Answer:

(a) $0.690$ s
(b) $6.78$ m/s