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A car is traveling at a constant velocity of magnitude v0 when the driver notices a garbage can on the road in front of him. at that moment, the distance between the garbage can and the front of the car is
d. a time t after noticing the garbage can, the driver applies the brakes and slows down at a constant rate before coming to a halt just before the garbage can. what is the magnitude of the car's acceleration after the brakes are applied?

Respuesta :

Answer:

[tex]a = \frac{v_o^2}{2(d - v_ot)}[/tex]

Explanation:

distance moved by the car while the driver reacted to apply brakes is given as

[tex]d_1 = v_o t[/tex]

now the distance of the car from the position of garbage is given as

[tex]d_2 = d - v_o t[/tex]

now if driver applies rakes due to which car decelerates uniformly and comes to rest

so we have

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

[tex]0 - v_o^2 = 2(-a)(d - v_ot)[/tex]

[tex]a = \frac{v_o^2}{2(d - v_ot)}[/tex]

onyebuchinnaji

The magnitude of the acceleration of the car is [tex]a = \frac{v_0^2}{2d- 2v_0t}[/tex]

The given parameters;

  • the constant velocity of the car, = [tex]v_0[/tex]
  • the distance between the car and the garbage = d
  • the time when the driver noticed the garbage = t

To find:

  • the magnitude of the car's acceleration after the brakes have been applied

The distance traveled by the  car  before the driver discovers the garbage;

[tex]d_1 = v_0t[/tex]

The new distance between the driver and garbage;

[tex]d_2 = d - v_0t[/tex]

The magnitude of the acceleration of the car is calculated as;

[tex]v_f^2 = v_0^2 + 2ad_2[/tex]

when the driver reaches the final distance ([tex]d_2[/tex]) after the breaks are  applied, the final velocity ([tex]v_f[/tex]) of the car will be zero.

[tex]0 = v_0^2 + 2ad_2\\\\2ad_2 = v_0^2\\\\a= \frac{v_0^2}{2d_2}\\\\recall, \ d_2 = d- v_0t\\\\a = \frac{v_0^2}{2d - 2v_0t}[/tex]

Thus, the magnitude of the acceleration of the car is [tex]a = \frac{v_0^2}{2d- 2v_0t}[/tex]

Learn more here: https://brainly.com/question/18369240

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