12) The velocity of a car decreased uniformly from \( 54 \mathrm{~km} / \mathrm{hr} \) to \( 36 \mathrm{~km} / \mathrm{hr} \). If the covered distance 25 m . during that interval , then the acceleration of the car \( =\ldots \ldots . . . \mathrm{m} / \mathrm{sec}^{2} \) \( \begin{array}{llll}\text { (a) } \frac{2}{5} & \text { S? }\end{array} \)

To find the acceleration of the car, we'll first convert the initial and final velocities from kilometers per hour to meters per second. 1. The initial velocity (\( u \)) is \( 54 \mathrm{~km/hr} = 54 \times \frac{1000}{3600} = 15 \mathrm{~m/s} \). 2. The final velocity (\( v \)) is \( 36 \mathrm{~km/hr} = 36 \times \frac{1000}{3600} = 10 \mathrm{~m/s} \). Using the kinematic equation \( v^2 = u^2 + 2as \), where \( s \) is the distance covered (25 m), we can solve for acceleration (\( a \)): \[ 10^2 = 15^2 + 2a(25) \\ 100 = 225 + 50a \\ 50a = 100 - 225 \\ 50a = -125 \\ a = -\frac{125}{50} = -2.5 \, \text{m/s}^2 \] So the acceleration of the car is \( -2.5 \, \text{m/s}^2 \), indicating a deceleration. Acceleration \( = -2.5 \, \text{m/s}^2 \)