Answer:
He has a speed of 16.60m/s after 35.0 meters.
Explanation:
The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:
[tex]v_{f}^{2} = v_{i}^{2} + 2ad[/tex]
[tex]v_{f} = \sqrt{v_{i}^{2} + 2ad}[/tex] (1)
The acceleration can be found by means of Newton's second law:
[tex]\sum F_{net} = ma[/tex]
Where [tex]\sum F_{net}[/tex] is the net force, m is the mass and a is the acceleration.
[tex]Fx + Fy = ma[/tex] (2)
All the forces can be easily represented in a free body diagram, as it is shown below.
Forces in the x axis:
[tex]F_{x} = F - F_{air}[/tex] (3)
Forces in the y axis:
[tex]F_{y} = 0[/tex] (4)
Solving for the forces in the x axis:
[tex]F_{x} = F - F_{air}[/tex]
Where [tex]F = 1.150x10^{3} N[/tex] and [tex]F_{air} = 795 N[/tex]:
[tex]F_{x} = 1.150x10^{3} N - 795 N[/tex]
[tex]F_{x} = 355 N[/tex]
Replacing in equation (2) it is gotten:
[tex]Fx + Fy = ma[/tex]
[tex]355 N + 0 N = (90.0 Kg)a[/tex]
[tex]355 N = (90.0 Kg)a[/tex]
[tex]a = \frac{355 N}{90.0Kg}[/tex]
[tex]a = \frac{355 Kg.m/s^{2}}{90.0Kg}[/tex]
[tex]a = 3.94 m/s^{2}[/tex]
So the acceleration for the cyclist is [tex]3.94 m/s^{2}[/tex], now that the acceleration is known, equation (1) can be used:
[tex]v_{f} = \sqrt{v_{i}^{2} + 2ad}[/tex]
However, since he was originally at rest its initial velocity will be zero ([tex]v_{i} = 0[/tex]).
[tex]v_{f} = \sqrt{2ad}[/tex]
[tex]v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}[/tex]
[tex]v_{f} = 16.60m/s[/tex]
He has a speed of 16.60m/s after 35.0 meters
