Entrance and exit ramps for freeways in (Figure 1) are often circular stretches of road. As you go around one at a constant spee

Question

3 years ago

8

Entrance and exit ramps for freeways in (Figure 1) are often circular stretches of road. As you go around one at a constant spee

d, you will experience a constant acceleration. Suppose you drive through an entrance ramp at a modest speed and your acceleration is 3.0 m/s2 .What will be the acceleration if you triple your speed?

Physics

2 answers:

LekaFEV [45] · Answer 1 · 2022-08-29T03:10:40+03:00

Answer:

27 m/s²

Explanation:

Traveling in uniform circular motion will cause a centripetal acceleration, an acceleration towards the center of the circle.

The acceleration of an object in circular motion is defined as a = v²/r.

Let's set the radius to be 5 m, since it is always unchanging. Solving for the velocity using a = 3 m/s² and r = 3 m gives us:

3 = v²/3
9 = v²
3 = v

The velocity is 3 m/s when the acceleration is 3 m/s² and the radius of the circular path is 3 m.

Now let's triple the velocity: 3 m/s → 9 m/s.

Let's solve for the acceleration using v = 9 m/s and r = 3 m.

a = (9)²/3
a = 81/3
a = 27

The acceleration is 27 m/s² if you triple your speed.

This makes sense because a is directly proportional to v², meaning that if you triple your speed, you increase the acceleration a by a factor of 3², which is a 9 times increase.

a = 3 * 9 = 27 m/s²

As a sidenote, doubling your speed would increase the acceleration by a factor of 2² - a 4 times increase.

wolverine [178] · Answer 2 · 2022-08-29T05:36:45+03:00

Answer:

27 m/s²

Explanation:

Traveling in uniform circular motion will cause a centripetal acceleration, an acceleration towards the center of the circle.

The acceleration of an object in circular motion is defined as a = v²/r.

Let's set the radius to be 5 m, since it is always unchanging. Solving for the velocity using a = 3 m/s² and r = 3 m gives us:

3 = v²/3

9 = v²

3 = v

The velocity is 3 m/s when the acceleration is 3 m/s² and the radius of the circular path is 3 m.

Now let's triple the velocity: 3 m/s → 9 m/s.

Let's solve for the acceleration using v = 9 m/s and r = 3 m.

a = (9)²/3

a = 81/3

a = 27

The acceleration is 27 m/s² if you triple your speed.

This makes sense because a is directly proportional to v², meaning that if you triple your speed, you increase the acceleration a by a factor of 3², which is a 9 times increase.

a = 3 * 9 = 27 m/s²

As a sidenote, doubling your speed would increase the acceleration by a factor of 2² - a 4 times increase.