Answer:
Time, t = 0.23 seconds
Explanation:
It is given that,
Initial speed of the ranger, u = 52 km/h = 14.44 m/s
Final speed of the ranger, v = 0 (as brakes are applied)
Acceleration of the ranger, 
Distance between deer and the vehicle, d = 87 m
Let d' is the distance covered by the deer so that it comes top rest. So,


d' = 26.06 m
Distance between the point where the deer stops and the vehicle is :
D=d-d'
D=87 - 26.06 = 60.94 m
Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :


t = 0.23 seconds
Hence, this is the required solution.
Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Work done = force * distance moved (in direction of the force)
force= mass* acceleration
force=58.1N
58.1*(5.8*10^4)
=3,369,800 J
Explanation:
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