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To solve for the radius \( r \) of the loop, we can use the centripetal acceleration formula:

\[
a = \frac{v^2}{r}
\]

where:
- \( a \) is the centripetal acceleration,
- \( v \) is the velocity,
- \( r \) is the radius of the loop.

Rearrange this formula to solve for \( r \):

\[
r = \frac{v^2}{a}
\]

Given:
- \( v = 14.2 \, \text{m/s} \)
- \( a = 13.6 \, \text{m/s}^2 \)

Substitute these values into the formula:

\[
r = \frac{(14.2)^2}{13.6}
\]

\[
r = \frac{201.64}{13.6}
\]

\[
r \approx 14.83 \, \text{m}
\]

### Answer:
The radius of the loop is approximately \( 14.83 \, \text{m} \).

Would you like further explanation or have any questions?

Here are some related questions to explore this topic further:
1. How does centripetal force relate to centripetal acceleration and radius?
2. What would happen to the radius if the velocity doubled but acceleration remained the same?
3. How would you calculate the centripetal force if the mass of the rollercoaster was known?
4. What is the importance of centripetal acceleration in circular motion?
5. How does the radius affect the speed required for an object to stay in circular motion?

**Tip:** Remember that centripetal acceleration always points towards the center of the circular path, keeping the object moving in a circular direction.

A rollercoaster travels through a loop at 14.2 m/s, and has a centripetal acceleration of 13.6 m/s². What is the radius of the loop?

Find out how to calculate the radius of a rollercoaster loop using velocity and centripetal acceleration. In this example, the rollercoaster moves at 14.2 m/s with a centripetal acceleration of 13.6 m/s². Suitable for high school physics students studying circular motion.

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