REALLY HARD Physics/Math Question!!!?

Question

A ranger in a national park is driving at 45 km/h when a deer jumps onto the road 64 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of -3.40 m/s2. What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?

find out seconds for reaction

Please answer its really hard.. I don't get it T.T Thanks!!!

Answer 1

hmm., are you sure person above me?

Answer 2

reaction time (t) is that time during which driver kept moving with v=45 km/h

v = 45*5/18 = 12.5 m/s

distance covered duting (t), s1 = 12.5 *t

now breaks applied, speed will go down

remaining distance (s) = 64 - s1 = 64 - 12.5 t

v(final)^2 = v^2 + 2 a s

v(final)^2 = 12.5^2 - 2*3.4 s

v(final)^2 = v^2 + 2 a [64 - 12.5 t]

stops, v(final)=0

v(final)^2 = 12.5^2 - 6.8*[64 - 12.5 t]

0 =  12.5^2 - 6.8*[64 - 12.5 t]

64 -12.5t =  [12.5*12.5/6.8] = 2.3 (approx)

t = 3.282 s

3.28 s

t = 3.25 s will be safer

he had 5.12 s = 64/12.5 if not reacted at all

--------------------------------------...

you can do by v-t graph as well

s = area of graph = area of rectangle + area of triangle

in triangle >> height =12.5

base = time left = (0 - 12.5)/a = - 12.5 / -3.4 = 12.5/3.4

64 = 12.5 * t + (1/2) * 12.5 [12.5/3.4]

Answer 3

Let t = maximum reaction time

For the reaction time to be maximum so that the ranger is to avoid hitting the deer,  the distance he traveled from the moment he sees the deer to the moment he reaches the deer must be equal to 64 m.

Now,

64 m = Distance traveled during reaction time + Distance traveled when decelerating at - 3.40 m/s^2

64 m = Vi(t) + [Vi(t) + (1/2)(a)(t^2)]

64m =  (45km/hr)t + [(45km/hr)(t) + (1/2)(a)(t^2)]

64 = 45t + 45t + (1/2)(-3.40)(t^2)

64 = 90t - 1.7t^2

Reaarange...

1.7t^2 - 90t + 64 = 0

Solving the quadratic equation by the quadratic formula...

t = {- (-90) +/- sqrt[(-90)^2 - 4(1.7)(64)]}/2(1.7)

t = 0.72 sec   ANSWER

Hope this helps.

teddy boy

Answer 4

2 seconds