**A cyclist accelerates from rest at a rate of 1.00 .**

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## Question “A cyclist accelerates from rest at a rate of 1.00 .”

A cyclistaccelerates from rest at a rate of 1.00 .

How fast will a point on the rim of the tire ( = 68 ) at the top be moving after 3.7 ? [Hint: At any moment, the lowest point on the tire is in contact with theground and is at rest – see the figure .]

## Answer

### Answer 1

7

### Answer 2

Velocity at the tire’s center after 2.6 seconds = 0 + 2/6 = 2.6 m/sec

The distance between the top and the center of the tire is twice that of its diameter.

As v = wr

The angular velocity for the tire w is equal.

The distance from the top point to the center of the tire is twice as far as the distance from ther center.

Velocity at the top will be twice that **velocity = 2*2.6 = 5 m/sec**

### Answer 3

A cyclist accelerates from rest at a rate of 1.00 m/s2

${\mathrm{m}/\mathrm{s}}^{2}$. How fast will a point at the top of the rim of the tire (diameter = 68 cm

$\mathrm{c}\mathrm{m}$) be moving after 3.0 s?

## Conclusion

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