A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 99m?

To determine how long it will take for the car to come to a stop, we can use the equations of motion. Let's define the variables: - Initial velocity, \( u = 21.8 \, \text{m/s} \) - Final velocity, \( v = 0 \, \text{m/s} \) (the car eventually stops) - Displacement, \( s = 99 \, \text{m} \) First, we can use the following kinematic equation that relates these variables: \[ v^2 = u^2 + 2as \] Where: - \( a \) is the acceleration (which will be negative since the car is decelerating). Substituting in the known values: \[ 0 = (21.8)^2 + 2a(99) \] Calculating \( 21.8^2 \): \[ 21.8^2 = 475.24 \] So the equation becomes: \[ 0 = 475.24 + 198a \] Rearranging gives: \[ 198a = -475.24 \] Solving for \( a \): \[ a = \frac{-475.24}{198} \approx -2.40 \, \text{m/s}^2 \] Now we can find the time \( t \) it takes to stop using the first equation of motion: \[ v = u + at \] Substituting in our known values: \[ 0 = 21.8 + (-2.40)t \] Solving for \( t \): \[ 2.40t = 21.8 \] \[ t = \frac{21.8}{2.40} \approx 9.08 \, \text{s} \] Thus, it will take approximately **9.08 seconds** for the car to accelerate uniformly to a stop over the distance of 99 meters.