Answer:
I. Stopping acceleration = 6 m/s²
II. Stopping distance, S = 75 meters
Explanation:
Given the following data;
Final velocity = 30 m/s
Time = 5 seconds
To find the stopping acceleration;
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation;
[tex]a = \frac{30 - 0}{5}[/tex]
[tex]a = \frac{30}{5}[/tex]
Acceleration = 6 m/s²
II. To find the stopping distance, we would use the third equation of motion;
[tex] V^{2} = U^{2} + 2aS [/tex]
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
30² = 0² + 2*6*S
900 = 12S
S = 900/12
S = 75 meters
