Where \( s \) is the distance traveled by an object (in meters) with an initial velocity \( u(\mathrm{in} \mathrm{m} / \mathrm{s} \) ) and final velocity \( v \) (in \( \mathrm{m} / \mathrm{s} \) ), \( t \) seconds later. Find \( s \) when \( u \) is \( 8 \mathrm{~m} / \mathrm{s}, v \) is \( 12 \mathrm{~m} / \mathrm{s} \), and \( t \) is 5 seconds. A. 50 m B. 156 m C. 78 m D. 100 m

To find the distance \( s \) traveled by the object, we can use the formula for distance when initial velocity \( u \), final velocity \( v \), and time \( t \) are known: \[ s = ut + \frac{1}{2}(v - u)t \] We can also use the average velocity method, where the average velocity \( v_{avg} \) is given by: \[ v_{avg} = \frac{u + v}{2} \] Then, the distance \( s \) can be calculated as: \[ s = v_{avg} \cdot t \] Let's calculate it using both methods. ### Method 1: Using the first formula 1. Given: - \( u = 8 \, \text{m/s} \) - \( v = 12 \, \text{m/s} \) - \( t = 5 \, \text{s} \) 2. Calculate the distance \( s \): \[ s = ut + \frac{1}{2}(v - u)t \] Substituting the values: \[ s = 8 \cdot 5 + \frac{1}{2}(12 - 8) \cdot 5 \] Calculating each part: \[ s = 40 + \frac{1}{2}(4) \cdot 5 \] \[ s = 40 + 2 \cdot 5 \] \[ s = 40 + 10 \] \[ s = 50 \, \text{m} \] ### Method 2: Using average velocity 1. Calculate the average velocity: \[ v_{avg} = \frac{u + v}{2} = \frac{8 + 12}{2} = \frac{20}{2} = 10 \, \text{m/s} \] 2. Calculate the distance \( s \): \[ s = v_{avg} \cdot t = 10 \cdot 5 = 50 \, \text{m} \] Both methods give the same result. Thus, the distance \( s \) traveled by the object is: \[ \boxed{50 \, \text{m}} \]