Answer:
a = 0.21 m/s²
Explanation:
The average speed measured by the strip system is:
[tex]v=\frac{110m}{T}\\T = \text{time elapsed betwen first and second strip}[/tex]
We need time T to be such that v < (21m/s). So we must find the acceleration a such that:
[tex]v=\frac{110}{T}=21\\\\T=\frac{110}{21}\\[/tex]
We are dropping units for simplicity. Time is always seconds and distance meters.
Lets call x to car position, being x=0 at the first strip. The movement with constant acceleration is:
[tex]x=33t-\frac{a}{2} t^2[/tex]
We take a negative acceleration because the car must be slowing down.
At time t = 110/21 we need x = 110:
[tex]110=33\frac{110}{21}-\frac{a}{2} (\frac{110}{21})^2\\ a = 0.21[/tex]
So, the minimum acceleration is a = 0.21 m/s²
