The cyclist accelerates for 4 seconds.
Explanation:
The motion of the cyclist is a motion at constant acceleration, so we can use suvat equations:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 64 m is the distance covered
u = 15 m/s is the initial velocity
t is the time
[tex]a=0.50 m/s^2[/tex] is the acceleration
Substituting into the equation,
[tex]64=15t+\frac{1}{2}(0.50)t^2\\0.25t^2+15t-64=0[/tex]
Solving the equation for t,
[tex]t=\frac{-15\pm \sqrt{(15)^2-4(0.25)(-64)}}{2(0.25)}[/tex]
Which gives two solution: one is negative, so we discarde it since it has no physical meaning; the other one is
t = 4 s
which is the answer we were looking for.
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