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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.25 m/s2 for 4.15 s, making straight skid marks 62.5 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? You may want to calculate the initial velocity of the car first. m/s (b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2

Respuesta :

Answer:

Part a)

Final speed of the car is

[tex]v_f = 4.17 m/s[/tex]

Part b)

Acceleration of the car is

[tex]a = -5.39 m/s^2[/tex]

Explanation:

As we know that car makes a skid of 62.5 m

here acceleration of the car is

[tex]a = - 5.25 m/s^2[/tex]

now we have

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]62.5 = v_i (4.15) + \frac{1}{2}(-5.25)(4.15^2)[/tex]

[tex]v_i = 25.95 m/s[/tex]

Part a)

Speed of the car by which it will hit the tree

[tex]v_f = v_i + at[/tex]

[tex]v_f = 25.95 - (5.25)(4.15)[/tex]

[tex]v_f = 4.17 m/s[/tex]

Part b)

Now if car will stop after travelling same distance which same initial speed

Then we can use kinematics

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 25.95^2 = 2(a)(62.5)[/tex]

[tex]a = -5.39 m/s^2[/tex]

Acceleration of a object is the rate of change of velocity of the object per unit time.

  • a) The speed does the car then strike the tree 4.17 m/s.
  • b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.

What is the acceleration of a object?

Acceleration of a object is the rate of change of velocity of the object per unit time.

Given information-

The car slows uniformly with acceleration of −5.25 m/s squared for 4.15 s.

The skid made by the car is 62.5 m long.

The initial velocity of the car can be find out using the distance formula of motion as,

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Put the values as,

[tex]62.5=u\times4.15+\dfrac{1}{2}\times(-5.25)\times(4.15)^2\\u=25.95[/tex]

Thus the value of initial velocity is 25.95 m/s.

  • (a) The speed does the car then strike the tree-

The velocity formula using the equation of motion can be given as,

[tex]v=u+at\\v=25.95+(-5.25)\times4.15\\v=4.17\rm m/s[/tex]

Thus, the speed does the car then strike the tree 4.17 m/s.

  • (b) The acceleration need to be, so that the car narrowly avoids a collision-

To stop the car the final velocity of it must be 0. Thus,

[tex]2ad=v^2-u^2\\2\times a\times62.5=0-25.95^2\\a=-5.93\rm m/s^2[/tex]

The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.

Hence,

  • a) The speed does the car then strike the tree 4.17 m/s.
  • b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.

Learn more about the acceleration here;

https://brainly.com/question/605631

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