Answer:
The stopping distance is 36.15m
Explanation:
In this exercise we are going to apply the second equation of motion to model the situation
[tex]s=ut-\frac{1}{2}at^2[/tex]
Given data
time in seconds= 4.6 seconds
speed v= 11 m/s
deacceleration a= 8.2m/s
Substituting this values into the expression for distance we have
[tex]s=11*4.6-\frac{1}{2}*8.2*4.6^2 \\\\s=50.6-\frac{1}{2}*8.2*21.16 \\\\s=50.6-\frac{1}{2}*173.512 \\\\\s=50.6-86.756 \\\\s=36.15m[/tex]
The stopping distance is 36.15m
