# Find the speed of skateboarder as he reaches D. A skating track consists of a horizontal part AB and a curved part BC. The curved part can be modelled as a smooth semi-circular arc with centre O...

Find the speed of skateboarder as he reaches D.

A skating track consists of a horizontal part AB and a curved part BC. The curved part can be modelled as a smooth semi-circular arc with centre O and radius 2m, with C vertically above B. A skateboarder on the track may be modelled as a moving particle. He has a speed of 9.5m/s as he reaches the point B. The point D is on the arc BC and OD makes an angle 60 degrees with the upward vertical.

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We calculate this by considering the change in the potential energy and kinetic energy from point B to point D.

By the law of the conservation of energy PE + KE remains unchanged.

At B, PE = 0 as the skater has nowhere to fall and KE = 1/2mv1^2, where m is the mass of the skater and v1 is the speed at B = 9.5m/s.

At D, PE = mgh, where g is the acceleration due to gravity and h is the height of D above the ground, and KE = 1/2mv2^2, where v2 is the speed at D which we wish to calculate.

Now, the angle the line OD makes with the horizontal is 30 degrees, so

h = rsin30 + r = 2sin30 + 2 = 3.

Therefore,

1/2mv2^2 + 3mg = 1/2mv1^2

implies v2^2 + 6g = v1^2

implies v2 = sqrt(v1^2 - 6g) = sqrt(9.5^2 - 6(9.8)) = 5.6 m/s

**The speed at D is 5.6m/s**