A race car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s in 11s, moving on a circular track of radius 500 m.

Determine the tangential and centripetal components of the net force excreted on the car(by the ground) when it's speed is 15 m/s. the cars mass is 1100 kg.

Tangential force = (tangential acceleration) x (mass)

= (35/11)m/s^2*1100 = 3500 N

Centripetal force at 15 m/s = (V^2/R)*(mass)
= 15^2/500 x 1100 = 495 N

(It continues to rise as the car accelerates)

Well, well, well, looks like our little race car is in need of some force analysis! Alright, let's get tangential and centripetal, shall we?

First, let's calculate the tangential component of the net force. The tangential force is responsible for the car's acceleration along the circular path. We know the car's mass is 1100 kg, its initial velocity is 0 m/s, and its final velocity is 15 m/s. Using Newton's second law of motion (F = ma), we can find the tangential force:

F_tangential = m * (v_final - v_initial) / t

F_tangential = 1100 kg * (15 m/s - 0 m/s) / 11 s

F_tangential ≈ 1500 N

Alright, that takes care of the tangential component. Now onto the centripetal component of the net force. The centripetal force is directed towards the center of the circular path and keeps the car moving in the circle. It can be found using the formula:

F_centripetal = m * v^2 / r

F_centripetal = 1100 kg * (15 m/s)^2 / 500 m

F_centripetal ≈ 495 N

So, the tangential component of the net force exerted on the car (by the ground) when its speed is 15 m/s is approximately 1500 N, while the centripetal component is around 495 N.

And now, we have a forceful answer for our speedy race car! Keep those wheels turning, my friend!

To determine the tangential and centripetal components of the net force exerted on the car when its speed is 15 m/s, we need to first find the acceleration of the car.

Given:
Speed at t=0 (u) = 0 m/s
Speed at t=11s (v) = 35 m/s
Radius of the circular track (r) = 500 m
Mass of the car (m) = 1100 kg

We can use the following kinematic equation to find the acceleration (a) of the car:

v = u + at

Substituting the given values into the equation, we can solve for the acceleration:

35 = 0 + a * 11
35 = 11a
a = 35/11
a ≈ 3.18 m/s²

Now, we can calculate the tangential and centripetal components of the net force on the car when its speed is 15 m/s.

Given:
Speed of the car (v) = 15 m/s

Tangential Component of Force (Ft):
The tangential component of force is responsible for changing the speed of an object. It can be calculated using the equation:

Ft = m * a

Substituting the values:

Ft = 1100 * 3.18
Ft ≈ 3498 N

Centripetal Component of Force (Fc):
The centripetal component of force is responsible for maintaining the circular motion of an object. It can be calculated using the equation:

Fc = m * v² / r

Substituting the values:

Fc = (1100 * 15²) / 500
Fc ≈ 495 N

Therefore, when the car's speed is 15 m/s, the tangential component of the net force exerted on the car by the ground is approximately 3498 N, and the centripetal component of the net force is approximately 495 N.

To find the tangential and centripetal components of the net force exerted on the car when its speed is 15 m/s, we can start by calculating the acceleration of the car using the given information.

The car starts from rest and reaches a speed of 35 m/s in 11 seconds. Using the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can rearrange the equation to solve for the acceleration:

a = (v - u) / t

Substituting the given values:

a = (35 m/s - 0 m/s) / 11 s
a = 3.18 m/s²

Now that we have the acceleration, we can calculate the tangential and centripetal forces.

1. Tangential Component of the Net Force:
The tangential component of the net force is responsible for changing the speed of the car.

Ft = m * a

where Ft is the tangential force, m is the mass of the car, and a is the acceleration.

Substituting the given values:

Ft = 1100 kg * 3.18 m/s²
Ft = 3498 N (rounded to the nearest whole number)

2. Centripetal Component of the Net Force:
The centripetal component of the net force is responsible for keeping the car moving in a circular path.

Fc = (m * v²) / r

where Fc is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the circular track.

Substituting the given values:

Fc = (1100 kg * (15 m/s)²) / 500 m
Fc = 495 N (rounded to the nearest whole number)

Therefore, when the car's speed is 15 m/s, the tangential component of the net force exerted by the ground on the car is 3498 N, and the centripetal component of the net force is 495 N.