Eond: Essay body moves in a straight line with an initial velocity of \( 7 \mathrm{~m} / \mathrm{s} \) and a constant acceleration \( \mathrm{m} / \mathrm{s}^{2} \) in the same direction. Find: e final velocity and the distance traveled after 6 seconds.

To find the final velocity and distance traveled by the object, we can use the equations of motion. 1. **Final Velocity**: The formula to calculate the final velocity \( v \) is: \( v = u + at \) where: - \( u = 7 \, \text{m/s} \) (initial velocity) - \( a = 2 \, \text{m/s}^2 \) (assuming a value for constant acceleration, since it was not specified) - \( t = 6 \, \text{s} \) Plugging in the values: \( v = 7 + (2 \times 6) = 7 + 12 = 19 \, \text{m/s} \) 2. **Distance Traveled**: The formula for distance \( s \) is: \( s = ut + \frac{1}{2}at^2 \) Using the same values: \( s = 7 \times 6 + \frac{1}{2} \times 2 \times (6^2) \) \( s = 42 + \frac{1}{2} \times 2 \times 36 \) \( s = 42 + 36 = 78 \, \text{m} \) So, after 6 seconds, the final velocity is \( 19 \, \text{m/s} \) and the distance traveled is \( 78 \, \text{m} \).