2. A motor boat starts from rest on a lake and accelerates in a straight line at a constant rate of 3.0ms^-1 for 8s. What distance does the boat cover during this time? 3. A boy standing on a platform throws a ball straight up with an initial velocity of 11.2ms^-1

2. The distance covered by the boat can be calculated using the equation of motion: <br /><br />\[ \text{Distance} = \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 \]<br /><br />Given that the acceleration is \(3.0 \, \text{m/s}^2\) and the time is \(8 \, \text{s}\), we can substitute these values into the equation:<br /><br />\[ \text{Distance} = \frac{1}{2} \times 3.0 \, \text{m/s}^2 \times (8 \, \text{s})^2 \]<br /><br />\[ \text{Distance} = \frac{1}{2} \times 3.0 \, \text{m/s}^2 \times 64 \, \text{s}^2 \]<br /><br />\[ \text{Distance} = 96 \, \text{m} \]<br /><br />So, the boat covers a distance of 96 meters during this time.<br /><br />3. The maximum height reached by the ball can be calculated using the equation of motion:<br /><br />\[ \text{Maximum Height} = \frac{\text{Initial Velocity}^2}{2 \times \text{Acceleration due to gravity}} \]<br /><br />Given that the initial velocity is \(11.2 \, \text{m/s}\) and the acceleration due to gravity is \(9.8 \, \text{m/s}^2\), we can substitute these values into the equation:<br /><br />\[ \text{Maximum Height} = \frac{(11.2 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} \]<br /><br />\[ \text{Maximum Height} = \frac{125.44 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} \]<br /><br />\[ \text{Maximum Height} = 6.4 \, \text{m} \]<br /><br />So, the maximum height reached by the ball is 6.4 meters.