Accelerated Motion Problem 2 A car was traveling along a roadway at a constant speed when the driver decided to accelerate to go around other traffic. The driver covered a distance of g9 m while accelerating at a constant rate. Figure out the acceleration of the car and the time the car was meolerating based on the speeds given below and the distance covered while accelerating Enter Answers Don't include units

To solve this problem, we can use the kinematic equations of motion. Given that the car was traveling at a constant speed before accelerating, we can assume the initial velocity (u) is 0 m/s. The distance covered during acceleration is given as 9 m. We need to find the acceleration (a) and the time (t) taken for this acceleration.<br /><br />Using the equation of motion:<br />\[ v^2 = u^2 + 2as \]<br /><br />Where:<br />- \( v \) is the final velocity<br />- \( u \) is the initial velocity (0 m/s)<br />- \( a \) is the acceleration<br />- \( s \) is the distance (9 m)<br /><br />Since \( u = 0 \):<br />\[ v^2 = 2as \]<br /><br />We need the final velocity \( v \) to proceed. Let's assume the final velocity \( v \) is given or can be measured. For example, if \( v = 18 \) m/s:<br /><br />\[ 18^2 = 2 \cdot a \cdot 9 \]<br />\[ 324 = 18a \]<br />\[ a = \frac{324}{18} \]<br />\[ a = 18 \, \text{m/s}^2 \]<br /><br />Now, to find the time \( t \) taken for this acceleration, we can use the equation:<br />\[ v = u + at \]<br /><br />Since \( u = 0 \):<br />\[ t = \frac{v}{a} \]<br /><br />Using \( v = 18 \) m/s and \( a = 18 \) m/s²:<br />\[ t = \frac{18}{18} \]<br />\[ t = 1 \, \text{second} \]<br /><br />So, the acceleration of the car is 18 m/s², and the time taken for this acceleration is 1 second.