How far will a motorcycle that accelerates at 1.28m/s^2 travel as its velocity increases from 10.0m/s to 20.0m/s 39.1 m 195 m 117 m 3.91 m

To solve this problem, we can use the kinematic equation:<br /><br />\[ v^2 = u^2 + 2as \]<br /><br />where:<br />- \( v \) is the final velocity,<br />- \( u \) is the initial velocity,<br />- \( a \) is the acceleration, and<br />- \( s \) is the distance traveled.<br /><br />Given:<br />- \( u = 10.0 \, \text{m/s} \)<br />- \( v = 20.0 \, \text{m/s} \)<br />- \( a = 1.28 \, \text{m/s}^2 \)<br /><br />We need to find \( s \).<br /><br />First, plug in the given values into the equation:<br /><br />\[ (20.0 \, \text{m/s})^2 = (10.0 \, \text{m/s})^2 + 2 \cdot 1.28 \, \text{m/s}^2 \cdot s \]<br /><br />\[ 400 \, \text{m}^2/\text{s}^2 = 100 \, \text{m}^2/\text{s}^2 + 2.56 \, \text{m/s}^2 \cdot s \]<br /><br />Next, solve for \( s \):<br /><br />\[ 400 \, \text{m}^2/\text{s}^2 - 100 \, \text{m}^2/\text{s}^2 = 2.56 \, \text{m/s}^2 \cdot s \]<br /><br />\[ 300 \, \text{m}^2/\text{s}^2 = 2.56 \, \text{m/s}^2 \cdot s \]<br /><br />\[ s = \frac{300 \, \text{m}^2/\text{s}^2}{2.56 \, \text{m/s}^2} \]<br /><br />\[ s \approx 117.19 \, \text{m} \]<br /><br />Therefore, the motorcycle will travel approximately 117 meters as its velocity increases from 10.0 m/s to 20.0 m/s.<br /><br />So, the correct answer is:<br /><br />\[ \boxed{117 \, \text{m}} \]