Q2. A car accelerates at a constant rate from 0 to \( 25 \mathrm{~m} / \mathrm{s} \) over a distance of 25 m . Approximately how long does it take the car to reach the velocity of \( 25 \mathrm{~m} / \mathrm{s} \) ? \( \begin{array}{llll}\text { A. } 1 \mathrm{~s} & \text { B. } 2 \mathrm{~s} & \text { C. } 4 \mathrm{~s} & \text { D. } 8 \mathrm{~s}\end{array} \)

To find the time it takes for the car to reach a velocity of \( 25 \, \text{m/s} \) while accelerating over a distance of \( 25 \, \text{m} \), we can use the kinematic equation: \[ v^2 = u^2 + 2a s \] where: - \( v \) is the final velocity (\( 25 \, \text{m/s} \)), - \( u \) is the initial velocity (\( 0 \, \text{m/s} \)), - \( a \) is the acceleration, - \( s \) is the distance (\( 25 \, \text{m} \)). Substituting the known values into the equation: \[ (25)^2 = (0)^2 + 2a(25) \] This simplifies to: \[ 625 = 50a \] Now, solving for \( a \): \[ a = \frac{625}{50} = 12.5 \, \text{m/s}^2 \] Next, we can use the formula for acceleration to find the time \( t \): \[ v = u + at \] Substituting the known values: \[ 25 = 0 + (12.5)t \] Solving for \( t \): \[ t = \frac{25}{12.5} = 2 \, \text{s} \] Thus, the time it takes for the car to reach a velocity of \( 25 \, \text{m/s} \) is approximately \( 2 \, \text{s} \). The answer is **B. 2 s**.