Question

A cyclist coasts up a 10.5° slope, traveling 19.0 m along the road to the top of the hill. If the cyclist's initial speed is 9.50 m/s, what is the final speed? Ignore friction and air resistance.

Answer 1

Answer:

Final speed, v = 12.57 m/s

Explanation:

Given that,

Slope with respect to horizontal, $\theta=10.5^{\circ}$

Distance travelled, d = 19 m

Initial speed of the cyclist, u = 9.5 m/s

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

$h=d\times sin\theta$

$h=19\times sin(10.5)$  

h = 3.46 m

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

$\dfrac{1}{2}m(v^2-u^2)=mgh$

$\dfrac{1}{2}(v^2-u^2)=gh$

$\dfrac{1}{2}\times (v^2-(9.5)^2)=9.8\times 3.46$

v = 12.57 m/s

So, the final speed of the 12.57 m/s. Hence, this is the required solution.