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Setler79 [48]
3 years ago
15

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 br

akes, 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.
Engineering
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

Explanation:

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

where;

v_o = initial velocity

a_c = constant deceleration

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

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

S = d + v_ot + \dfrac{1}{2}at^2

Then:

v_B = 60-12t

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

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

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

v_A = 60 - 18(t-0.75)

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

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

Relating both velocities:

v_B = v_A

60-12t = 60 - 18(t-0.75)

60-12t =73.5 - 18t

60- 73.5 = - 18t+ 12t

-13.5 =-6t

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.

S_B = S_A

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

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

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

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