Finding time using speed and acceleration?

Question

A ranger in a national park is driving at 54 km/h when a deer jumps onto the road 55 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of -3 m/s2. What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?

* i got 40 sec. but the answer was wrong.. i don't get it.

Answer 1

hello again. first u convert 54km/h to m/s which gives you 15m/s. then u find the time taken to decelerate from 15 to 0 which is 5 secs. (15m/s/3m/s^2). In 5 secs, the jeep can travel 37.5m/s, given by 0.5 x 15m/s x 5s. (remember the v-t graph thing in your previous question?)

so he has a distance of 55m-37.5m = 17.5m to react. okay so how long does it take him to travel 17.5 m? it's just distance/speed which is 17.5m/15m/s = 1.17s. there you go, he has 1.17 secs to react. come on, does anyone really take 40s to step on the brake pedal?

Answer 2

Lets put everything into m/s --> 54 km/hr = 15 m/s

The range travels the 55m in 3.67 seconds if he does nothing. Now if he comes to a stop after hitting the breaks, it has to be in less time than 3.67 seconds. SO lets see how long it takes to go from 15 m/s to zero when he hits the brakes.

Use

x = (v^2 - v0^2)/(2a) where v = 0, v0 = 15m/s and a = -3 m/s^2

x = -225m^2/s^2/(2*(-3m/s^2)) = 37.5 m

SO the jeep can travel 55 - 37.5 m = 17.5 m before hitting the brakes. The time it takes the jeep to travel this distance is

t = 17.5m/15m/s = 1.17 sec.

SO the range can wait 1.17 seconds before hitting the brake.

Answer 3

First convert 54 km/h to m/s = 15 m/s

Then it's formula time!

d = 1/2(vi + vf)t
55 = 1/2(15 + 0)t
t = 7.33 sec