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Question No. 1 Marks = = 5 +5 +2 = 12

The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.



Question No. 2 Marks = 8
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.

Sagot :

The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1.

     a) The stopping distance is: x = 270 m

     b) The initial velocity is: v₀ = 17.3 m / s

     c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

  • The stopping distance is: x = 4d

Given parameters

  • Mass of the red carriage m1 = 2,0 10³ kg
  • Red car speed vo = 45 m / s
  • Friction force fr = 7.5 10³ N.

To find

Question 1.

    a) Minimum braking distance.

    b) If the distance is x = 40.0 m, what speed should the vehicles have?

    c)  Conclusive importance of physics in daily life.

Question 2.

  • The distace to stop.

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

         v² = v₀² - a2 x

Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.

Newton's second law states that the net force is proportional to the mass and the acceleration of the body.

          F = ma

Where F is force, m is mass and acceleration.

In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body

        fr = m a

        a =[tex]\frac{fr}{m}[/tex]  

        a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]  

        a = 3.75 m / s²

This acceleration is in the opposite direction to the speed.

Let's find the distance needed to stop, the final speed is zero.

          0 = v₀² - 2 ax

           x = [tex]\frac{v_o^2 }{ 2a}[/tex]  

Let's calculate.

          x = [tex]\frac{45^2 }{2 3.75}[/tex]  

          x = 270 m

This is the minimum distance that the two vehicles must separate to avoid a collision.

b) We look for speed.

        v₀ = [tex]\sqrt{2ax}[/tex]  

        v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]  

        v₀ = 17.3 m / s

c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.

           0 = v₀² - 2ax

Let's substitute.

             a = [tex]\frac{v^2}{2d}[/tex]  

They ask the distance traveled if this car traveled from an initial speed 2v.

             0 = v² - 2 a x

             x = [tex]\frac{v^2}{2a}[/tex]  

We substitute

            x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]  

            x = 4 d

In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1

       a) The stopping distance is: x = 270 m

       b) The initial velocity is: v₀ = 17.3 m / s

        c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

  •  The stopping distance is x = 4d

Learn more about kinematics here:  brainly.com/question/13202578

View image Moya1316
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