the final speed of the cyclist is 32.23 miles per hour
given the data in the question;
initial velocity; u = 0 m/s
time; t = 4.30 seconds
Distance; S = 31.0 m
Final Velocity; u = ?
First we find the acceleration of the bicycle, using the second equation of motion;
[tex]S = ut + \frac{1}{2}at^2[/tex]
we substitute in our given values
[tex]31.0m = (0m/s * 4.30s) + ( \frac{1}{2} * a * ( 4.30s)^2 )\\[/tex]
[tex]31.0m = 0 + ( \frac{1}{2} * a * 18.49 s^2 )[/tex]
[tex]a = \frac{31.0 m}{9.245s^2}\\[/tex]
[tex]a = 3.35 m/s^2[/tex]
Now, we can easily find the final speed using the first equation of motion;
[tex]v = u + at[/tex]
so we substitute in our values
[tex]v = 0m/s + ( 3.35 m/s^2 * 4.30s )\\v = ( 3.35 m/s^2 * 4.30s )\\v = 14.41 m/s[/tex]
we know that; 1 m/s = 2.2369 miles per hour
Hence
[tex]v = 32.23 mph[/tex]
Therefore, the final speed of the cyclist is 32.23 miles per hour
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