Where $$ s $$ is the distance traveled by an object (in meters) with an initial velocity $$ u $$ (in $$ \mathrm{m} / \mathrm{s} $$ ) and final velocity $$ v $$ (in $$ \mathrm{m} / \mathrm{s} $$ ), seconds later. And $$ s $$ when $$ u $$ is $$ 9 \mathrm{~m} / \mathrm{s}, v $$ is $$ 13 \mathrm{~m} / \mathrm{s} $$, and $$ t $$ is 2 seconds. B. 44 m C. 22 m D. 15.5 m

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The distance $$ s $$ is calculated using the formula $$ s = ut + \frac{1}{2}(v - u)t $$. Given $$ u = 9 \, \text{m/s}, v = 13 \, \text{m/s}, t = 2 \, \text{s} $$, the distance $$ s $$ is $$ 22 \, \text{m} $$. Correct answer: C. 22 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: $$ s = ut + \frac{1}{2}(v - u)t $$ Given: - $$ u = 9 \, \text{m/s} $$ - $$ v = 13 \, \text{m/s} $$ - $$ t = 2 \, \text{s} $$ Now, we can calculate $$ s $$: 1. Calculate the first term $$ ut $$: $$ ut = 9 \, \text{m/s} \times 2 \, \text{s} = 18 \, \text{m} $$ 2. Calculate the change in velocity $$ v - u $$: $$ v - u = 13 \, \text{m/s} - 9 \, \text{m/s} = 4 \, \text{m/s} $$ 3. Calculate the second term $$ \frac{1}{2}(v - u)t $$: $$ \frac{1}{2}(v - u)t = \frac{1}{2} \times 4 \, \text{m/s} \times 2 \, \text{s} = \frac{1}{2} \times 8 \, \text{m} = 4 \, \text{m} $$ 4. Now, add both terms to find $$ s $$: $$ s = 18 \, \text{m} + 4 \, \text{m} = 22 \, \text{m} $$ Thus, the distance traveled $$ s $$ is $$ 22 \, \text{m} $$. The correct answer is **C. 22 m**.

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