6. A cart rolling down an incline has an acculeration of $$ 4.0 \mathrm{~m} / \mathrm{s}^{2} $$. If the cart has a beginning speed of 2.0 m / how long does it take to rench a speed of $$ 18 \mathrm{~m} / \mathrm{s} $$ ?

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It takes 4.0 seconds for the cart to reach a speed of 18 m/s. To find the time it takes for the cart to reach a speed of $$ 18 \, \text{m/s} $$ from an initial speed of $$ 2.0 \, \text{m/s} $$ with a constant acceleration of $$ 4.0 \, \text{m/s}^2 $$, we can use the following kinematic equation: $$ v = u + at $$ where: - $$ v $$ is the final velocity ($$ 18 \, \text{m/s} $$), - $$ u $$ is the initial velocity ($$ 2.0 \, \text{m/s} $$), - $$ a $$ is the acceleration ($$ 4.0 \, \text{m/s}^2 $$), - $$ t $$ is the time in seconds. Rearranging the equation to solve for $$ t $$: $$ t = \frac{v - u}{a} $$ Substituting the known values: $$ t = \frac{18 \, \text{m/s} - 2.0 \, \text{m/s}}{4.0 \, \text{m/s}^2} $$ $$ t = \frac{16 \, \text{m/s}}{4.0 \, \text{m/s}^2} $$ $$ t = 4.0 \, \text{s} $$ Therefore, it takes $$ 4.0 $$ seconds for the cart to reach a speed of $$ 18 \, \text{m/s} $$.

6. A cart rolling down an incline has an acculeration of \( 4.0 \mathrm{~m} / \mathrm{s}^{2} \). If the cart has a beginning speed of 2.0 m / how long does it take to rench a speed of \( 18 \mathrm{~m} / \mathrm{s} \) ?

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6. A cart rolling down an incline has an acculeration of \( 4.0 \mathrm{~m} / \mathrm{s}^{2} \). If the cart has a beginning speed of 2.0 m / how long does it take to rench a speed of \( 18 \mathrm{~m} / \mathrm{s} \) ?

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