Question

A cyclist coasts up a 9.00° slope, traveling 12.0 m along the road to the top of the hill. If the cyclist’s initial speed is 9.00 m/s, what is the final speed? Ignore friction and air resistance.

Answer 1

Answer:

$v_{f} \approx 6.647\,\frac{m}{s}$

Explanation:

The final speed of the cyclist is determined by applying the Principle of Energy Conservation:

$\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}$

$\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}$

$v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})$

$v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}$

$v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}$

$v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}$

$v_{f} \approx 6.647\,\frac{m}{s}$

Answer 2

Answer:

the final speed of the 10.85 m/s.

Explanation:

Given that,

Slope with respect to horizontal, $\theta=9^{\circ}$

Distance travelled, d = 12 m

Initial speed of the cyclist, u = 9 m/s

We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,

$h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m$

v is the final speed of the cyclist. It can be calculated using work energy theorem as

$\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s$

Thus,the final speed of the 10.85 m/s.