1. Not finding help here? Sign up for a free 30min 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!

Kinematics in two dimensions problem

  1. Jan 29, 2008 #1
    [SOLVED] Kinematics in two dimensions problem

    1. The problem statement, all variables and given/known data
    Chapter 3, Problem 20


    Incorrect.

    A car drives horizontally off the edge of a cliff that is 48.1 m high. The police at the scene of the accident note that the point of impact is 195 m from the base of the cliff. How fast was the car traveling when it drove off the cliff?


    2. Relevant equations

    y=V(0)t+ (1/2)gt^2

    y=(1/2)(V(0)+V)T- maybe?


    3. The attempt at a solution


    Ok I think I've figured out the first part of the equation, finding out the time it took for the car to fly off the cliff- here's how I did it.

    y= V(0)t+ (1/2)gt^2
    -48.1= (0)t +(1/2)(-9.8)t^2
    t=3.13

    Here's where I think I'm messing up somewhere:

    x=195 (how far the car flew off the cliff)
    t=3.13 (what i got from the previous answer)
    V(0)= ? (what i need to solve for)
    V= 0 ( the car doesn't have any more veloicty after it crashed into something)

    however, when i plug this into my equation :

    x=(1/2)(V(0)+V)t, it says i didn't get the right answer. here's my math just for a double check

    195=(1/2)(V(0)+0)(3.13)
    195=1.57V(0)
    V(0)= 124.5

    Any thoughts?
     
  2. jcsd
  3. Jan 29, 2008 #2


    Remember you have to treat the Y and the X completely different in this case. What would of happened to the car if there was no gravitiy pulling down from the cliff? Would it ever have been stopped by the ground? Would its velocity ever changed (disregarding air resistance). Gravity is only working here in the Y direction.
     
  4. Jan 29, 2008 #3
    I'm a little confused, wouldn't V=0 on the x-axis if the car had stopped/crashed, but on the Y-axis there still is the force of gravity. Or are you saying that my time i computed for the Y values won't work for my X values?
     
  5. Jan 29, 2008 #4
    No, the time traveled is the same for both the X and the Y. What I'm trying to convey is the point that the car crashed due to a constant acceleration downward in the Y direction (gravity). The motion of the car in the X direction is completely independent and without gravity would have continued traveling never hitting the ground. This is why using the Vf=0 in the X equation isn't valid (That would imply a negative acceleration which in turn would imply an additional force at work, which there isn't in the X direction). So, you found the travel time from the equations for the Y direction. This time we know has to be the same for the X direction. Also we know the distance the car traveled in the X direction, 195m. Can you figure for Vi from here? (remember there is nothing accelerating the car in the X direction)
     
  6. Jan 29, 2008 #5
    so it would be a logical assumption that g=0 in the x direction, correct?

    so an equaiton would look something like this?

    195=V(0)(3.13)+(1/2)(0)(3.13)^2
    195=3.13V(0)
    V(0)= 62.3 m/s?
     
  7. Jan 30, 2008 #6
    Looks good to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kinematics in two dimensions problem
Loading...