Question

A skater moves over ice in circular path at a constant speed. He later moves over ice in a circular path at a constant speed, but this time with five times as much acceleration as before.
What single difference in the motion of the skater might have caused his acceleration to be five times as great?

• A. He may have been moving five times as fast as before
• B. He may have been moving around a curve with five times the radius of the original circular path.
• C. He may have been moving a little over two times as fast
• D. He may have been moving around a curve with a radius just over twice as much as the original radius.
• E. he may have been wearing more massive skates.

Answer 1

Answer: (C)

He may have been moving a little over two times as fast

for objects moving at constant speed in circular paths, the only two factors that affect acceleration are speed and circular radius. the equation for this "centripetal" or "center-seeking" acceleration is:

$a_c=\frac{v^2}{r}$

We see from this equation that the acceleration for an object moving at a constant speed in a circular path is directly proportional to the square of the object's speed and inversely proportional to the radius of its circular path.

in this case, we want to make the acceleration 5 times as great.

since the acceleration of the skater is directly proportional to the square of the speed, making the speed 5 times as great will make the acceleration 25 times as great.

if we make the speed just over two times as great, we cause the acceleration to be just over "2 squared" times as great, or 5 times as great. (we would need to be more exact in our choice of speed to make the acceleration exactly 5 times as great).

The correct answer is choice C.