# Radial & Tangential Acceleration of a Car at Indianapolis 500 | Physics Solution

• y90x
In summary, the problem involves a car accelerating uniformly in a semicircular arc at the Indianapolis 500, reaching a speed of 320 km/h. The tangential acceleration is found to be 6.288 m/s^2 and the radial acceleration is found to be 39.5 m/s^2. However, the correct answer for the radial acceleration is 19.75 m/s^2. The mistake was due to using the final velocity at the end of the arc instead of the halfway point. Additionally, the coefficient of static friction needed for the car to maintain this acceleration on a flat curve with no slipping or skidding is being questioned.
y90x

## Homework Statement

A car at the Indianapolis 500 accelerates uniformly from the pit area, going from rest to 320km/h in a semicircular arc with a radius of 200 m. Determine the tangential and radial acceleration of the car when it is halfway through the arc, assuming constant tangential acceleration. If the curve were flat, what coefficient of static friction would be necessary between the tires and the road to provide this acceleration with no slipping or skidding?

The attempt at a solution
I solved for tagental acceleration :
At= vf^2- Vo^2 /2d
= 88.8^2/2(pi * 200)
=6.288 m/s^2

For the radial acceleration is :
Ac= v^2/r
=88.8^2/200
=39.5 m/s^2
Why am I getting it wrong (the radial acceleration)? :/
88.8 is in m/s btw (from the 320 km/h)

y90x said:
from the 320 km/h

y90x said:
from rest to 270 km h
?

haruspex said:
?

That first one was a typo , in the paper it says 320 km/h

y90x said:
when it is halfway through the arc
Did you read that correctly when you wrote:
y90x said:
Ac= v^2/r
=88.8^2/200
?

haruspex said:
Did you read that correctly when you wrote:

?

That would mean the 200 m changes to 100 m
And it’ll give me a higher number , 79.03 m/s^2

y90x said:
That would mean the 200 m changes to 100 m
No, the radius is 200m. But what velocity are you using in v2/r?

haruspex said:
No, the radius is 200m. But what velocity are you using in v2/r?

Ohh, that’s the final velocity when it’s at the end of the arc ? So I find the velocity when it’s halfway , then apply it in v^2/r

y90x said:
Ohh, that’s the final velocity when it’s at the end of the arc ? So I find the velocity when it’s halfway , then apply it in v^2/r
Yes.

## Related to Radial & Tangential Acceleration of a Car at Indianapolis 500 | Physics Solution

Radial acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is caused by the centripetal force.

## How is radial acceleration calculated?

The formula for radial acceleration is a = v^2/r, where a is the radial acceleration, v is the velocity of the object, and r is the radius of the circle.

## What is the difference between radial acceleration and tangential acceleration?

Radial acceleration is the acceleration towards the center of a circular path, while tangential acceleration is the acceleration along the tangent of the circle. They are perpendicular to each other and together make up the total acceleration of an object moving in a circular path.

## Why is radial acceleration important?

Radial acceleration is important because it is responsible for keeping objects in circular motion. Without it, objects would continue moving in a straight line instead of a curved path.

## How does radial acceleration relate to Newton's Laws of Motion?

Radial acceleration is a result of the centripetal force, which is required by Newton's First Law of Motion to keep an object moving in a circular path. It also plays a role in Newton's Second Law of Motion, as it is responsible for the change in velocity and direction of an object in circular motion.

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