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This problem involves the use of the kinematic equation for uniformly accelerated motion. The equation that relates the distance traveled \( x \), initial velocity \( u \), final velocity \( v \), and acceleration \( a \) is:

\[
v^2 = u^2 + 2ax
\]

### Given:
- Initial velocity, \( u = 0 \) m/s (since the cyclist starts from rest)
- Final velocity, \( v = 6 \) m/s
- Acceleration, \( a = 1.8 \) m/s\(^2\)

We need to find the distance \( x \) traveled by the cyclist.

### Applying the kinematic equation:

\[
v^2 = u^2 + 2ax
\]

Substituting the given values:

\[
(6)^2 = (0)^2 + 2(1.8)x
\]

\[
36 = 3.6x
\]

### Solving for \( x \):

\[
x = \frac{36}{3.6} = 10 \, \text{meters}
\]

### Conclusion:
The correct answer is **Option D: 10 m**.

Would you like more details or have any questions? Here are some related questions to consider:

1. What would the distance \( x \) be if the final velocity was 8 m/s instead of 6 m/s?
2. How would the distance change if the acceleration was reduced to 1 m/s²?
3. What would happen to the distance if the initial velocity was not zero?
4. Can you derive the kinematic equation used in this problem?
5. How is the motion affected if the cyclist experiences a deceleration instead of acceleration?

**Tip:** In uniformly accelerated motion, the distance covered is directly proportional to the square of the final velocity when starting from rest.

A cyclist starts from rest and accelerates uniformly down a slope. After covering a distance of x meters, the cyclist reaches a speed of 6 m/s. If the acceleration of the cyclist is 1.8 m/s², calculate the value of x.

This problem involves calculating the distance a cyclist covers while accelerating uniformly. Given a final velocity of 6 m/s and an acceleration of 1.8 m/s², find the distance traveled using kinematic equations. Suitable for students in Grades 9-11.

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