Aranger in a national park is driving at 52 km/h when a deer jumps onto the road 87 m ahead of the vehicle. after a reaction time of t s, the ranger applies the brakes to produce a deceleration of 4.0 m/s2 . what is the maximum reaction time allowed if the ranger is to avoid hitting the deer? answer in units of s.

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.

i need with 1.3.2 investigate cycling o2 and co2 apex

we need to see it

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j