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Correct Answer: after 3.4 seconds and 0.59 seconds
to find the time at which object a and object b will be in the same position, we need to first set up equations for the positions of both objects as functions of time.
let's denote the position of object a as \( x_a(t) \) and the position of object b as \( x_b(t) \), where \( t \) is the time in seconds after they start moving.
object a starts with a constant acceleration \( a = 5 \, \text{m/s}^2 \) from an initial position \( x_a(0) = 5 \, \text{m} \) (since it is 5 meters ahead). the initial velocity of object a is not given, so we assume it starts from rest, i.e., \( v_{a0} = 0 \, \text{m/s} \). using the kinematic equation for position under constant acceleration, the position of object a as a function of time is:
\[ x_a(t) = x_a(0) + v_{a0}t + \frac{1}{2}at^2 \]
\[ x_a(t) = 5 + 0 \cdot t + \frac{1}{2} \cdot 5 \cdot t^2 \]
\[ x_a(t) = 5 + 2.5t^2 \]
object b moves with a constant speed \( v_b = 10 \, \text{m/s} \) from the same starting point as a, which is \( x_b(0) = 0 \). since there is no acceleration, the position of object b as a function of time is simply:
\[ x_b(t) = x_b(0) + v_b \cdot t \]
\[ x_b(t) = 10t \]
to find when they are at the same position, we set \( x_a(t) \) equal to \( x_b(t) \):
\[ 5 + 2.5t^2 = 10t \]
rearranging the terms, we get a quadratic equation:
\[ 2.5t^2 - 10t + 5 = 0 \]
dividing the entire equation by 2.5 simplifies it to:
\[ t^2 - 4t + 2 = 0 \]
now, using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = 2 \), we find:
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]
\[ t = \frac{4 \pm \sqrt{16 - 8}}{2} \]
\[ t = \frac{4 \pm \sqrt{8}}{2} \]
\[ t = \frac{4 \pm 2\sqrt{2}}{2} \]
\[ t = 2 \pm \sqrt{2} \]
simplifying further, the solutions are:
\[ t_1 = 2 + \sqrt{2} \approx 3.41 \, \text{s} \]
\[ t_2 = 2 - \sqrt{2} \approx 0.59 \, \text{s} \]
therefore, objects a and b will be in the same position at approximately 0.59 seconds and 3.41 seconds after they start moving.
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WE