Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at ft/s^2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 18 ft/s^2.

Required:
Determine the minimum distance d between the cars so as to avoid a collision.

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2021-07-13T21:27:28+02:00

Answer:

Explanation:

Using the kinematics equation [tex]v = v_o + a_ct[/tex] to determine the velocity of car B.

where;

[tex]v_o =[/tex] initial velocity

[tex]a_c[/tex] = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

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

Then:

[tex]v_B = 60-12t[/tex]

The distance traveled by car B in the given time (t) is expressed as:

[tex]S_B = d + 60 t - \dfrac{1}{2}(12t^2)[/tex]

For car A, the needed time (t) to come to rest is:

[tex]v_A = 60 - 18(t-0.75)[/tex]

Also, the distance traveled by car A in the given time (t) is expressed as:

[tex]S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]

Relating both velocities:

[tex]v_B = v_A[/tex]

[tex]60-12t = 60 - 18(t-0.75)[/tex]

[tex]60-12t =73.5 - 18t[/tex]

[tex]60- 73.5 = - 18t+ 12t[/tex]

[tex]-13.5 =-6t[/tex]

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

[tex]S_B = S_A[/tex]

[tex]d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]

[tex]d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2[/tex]

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft