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An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3.35 m/s2 . if it takes him 0.200 s to get the brakes on and he is 20.0 m from the intersection when he sees the light, will he be able to stop in time?

Respuesta :

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

[tex]d_1 = v* t[/tex]

[tex]d_1 = 18* 0.200 = 3.6 m[/tex]

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

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

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

[tex]0^2 - 18^2 = 2*(-3.35)(s)[/tex]

[tex] s = 48.35 m[/tex]

So total distance after which car will stop is

[tex] d = d_1 + d_2[/tex]

[tex] d = 48.35 + 3.6 = 51.95 m[/tex]

So car will not stop before the intersection as it is at distance 20 m

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