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

the final speed of the 10.85 m/s.

Step-by-step 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* sin\thetah\\\\=12* sin(9) \\\\h = 1.877 m`

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

`(1)/(2)m(v^2-u^2)\\\\=mgh(1)/(2)(v^2-u^2)\\\\=gh(1)/(2)* (v^2-(9.0)^2)\\\\=9.8* 1.87\\\\v = 10.85 m/s`

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

Answer 2

`v_(f) \approx 6.647\,(m)/(s)`

Step-by-step explanation:

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

`(1)/(2)\cdot m\cdot v_(o)^(2) + m\cdot g\cdot h_(o) = (1)/(2)\cdot m\cdot v_(f)^(2) + m\cdot g\cdot h_(f)`

`(1)/(2) \cdot v_(o)^(2) + g\cdot (h_(o)-h_(f)) = (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\,(m)/(s) )^(2)-2\cdot (9.807\,(m)/(s) )\cdot (12\,m)\cdot \sin 9^(\textdegree)}`

`v_(f) \approx 6.647\,(m)/(s)`