# 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.