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Second: Essay A body moves in a straight line with an initial velocity of \( 7 \mathrm{~m} / \mathrm{s} \) and a constant acceleration of \( 4 \mathrm{~m} / \mathrm{s}^{2} \) in the same direction. Find: the final velocity and the distance traveled after 6 seconds.

Ask by Weaver Boyd. in Egypt
Mar 12,2025

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After 6 seconds, the final velocity is 31 m/s and the distance traveled is 114 meters.

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To find the final velocity, you can use the equation: \[ v = u + at \] where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Plugging in the values: \[ v = 7 \, \text{m/s} + (4 \, \text{m/s}^2 \times 6 \, \text{s}) = 7 \, \text{m/s} + 24 \, \text{m/s} = 31 \, \text{m/s} \] Now to find the distance traveled, use the equation: \[ s = ut + \frac{1}{2} a t^2 \] Substituting in the given values: \[ s = (7 \, \text{m/s} \times 6 \, \text{s}) + \frac{1}{2} (4 \, \text{m/s}^2 \times (6 \, \text{s})^2) = 42 \, \text{m} + \frac{1}{2} \times 4 \, \text{m/s}^2 \times 36 \, \text{s}^2 = 42 \, \text{m} + 72 \, \text{m} = 114 \, \text{m} \] So, the final velocity is \( 31 \, \text{m/s} \) and the distance traveled after 6 seconds is \( 114 \, \text{m} \).

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