1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two questions: Initial velocity and vehicle stopping distance

  1. Nov 9, 2011 #1
    Hi All,

    It's been a looong time since I've had to use/apply basic physics but i'm hoping i've come to the right place for help.

    I am trying to help my brother-in-law with a speeding charge.

    First question
    The police allege he reached 100km/h within 38m of a corner.

    The first question relates to the initial velocity a vehicle would have needed to be travelling in order to reach 100km/h (27.8 m/s) in 38m.

    I calculate that as this vehicle can accelerate from 0-100 km/h in 8.5, it would need 118.2m to get to 27.8 m/s from a standing start.

    But I am unsure how to calculate what the initial velocity would need to be for this vehicle to reach 27.8 m/s in 40m.

    Seconds question

    The police also allege he was still travelling at 27.8 m/s 47m from a speed hump. Assuming 1 sec reaction time, I calculate that he would have been travelling at:

    Formula used to calculate braking distance:

    Vf2 = V02 - 2ad

    where Vf is the final velocity, V0 is the initial velocity, a is the rate of deceleration and d is the distance travelled during deceleration. Since Vf will be zero when the car has stopped:

    d = v02 / 2a

    (I've assumed a = 10 m/s - is this realistic?)

    d = 772.8 / 20 = 38.6m

    Stopping distance incl. 1 sec reaction time = 38.6m + 27.8m = 66.4m

    So, velocity after 47m braking:

    Vf = Sq root (V02 - 2ad)

    = Sq root (772.8 -2 x 10 x 19.2)
    = 19.7 m/s
    = 71 km/h

    (where d = 47 metres minus the reaction distance of 27.8 metres = 19.2 metres)

    At 71 km/h he would have done some serious damage to his car.

    Thanks for your help - I apologise for my slopping logic in advance :smile:
  2. jcsd
  3. Nov 9, 2011 #2
    For your first problem, what would be the maximum speed the car could accelerate to from rest in 40m? :)
  4. Nov 9, 2011 #3


    User Avatar
    Homework Helper

    first question - Since the initial acceleration is greater than the final acceleration to reach 100 kph in 8.5 seconds, more time is spent at higher speed, so it would take more than 118 meters to accelerate from 0 to 100 kph. I'm not sure this matters, since 100 kph = 27.8 m/s, and at that speed traveling 38 meters would only take 1.37 seconds. Unless this is a very high powered car, acceleration from 80 kph to 100 kph would probably take more than 2 seconds and more than 50 meters. I'm assuming the corner could not be exited at 80 kph.

    second question - reaction time isn't a factor, assuming the driver could see the speed bump well before he needed to apply the brakes. An average hard braking deceleration rate would be around 7 m / s2.
    Last edited: Nov 9, 2011
  5. Nov 9, 2011 #4
    In the first question, you have everything correct, so far.

    Next, you need to calculate the car's maximum acceleration rate (based on 0-100 km/h in 8.5 sec). So, divide 27.8 m/s by 8.5 s to get 3.27 m/s/s.

    Now, use
    [tex] {V_f}^2 = {V_i}^2 + 2ad[/tex]
    to determine the initial velocity required to reach 100 km/h in 38 m.

    Plugging in your values, we get
    [tex](27.8\ m/s)^2 = {V_i}^2 + (2)(3.27\ m/s/s) (38\ m)[/tex]

    Solving for Vi, we get
    [tex]V_i = \sqrt{27.8^2 - (2) (3.27) (38)}\ \ m/s[/tex]
    [tex]V_i \approx 22.9\ m/s[/tex]
    Which works out to approximately 82.3 km/h

    He must have been screamin' around that corner!
  6. Nov 11, 2011 #5
    Thanks for the help guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook