1. A few problems on "MOTION IN ONE DIMENSION"

    INTRODUCTION: Here are a problem on "Motion in one-dimension".
    THE PROBLEMS: (1) "Car A is just behind Car B.Both are 5m long.The velocity of A is 15 m/s while that of B is 10 m/s. Find the road distnace covered by Car A to cross Car B."
    WHAT MY BRAIN SUGGESTED: Let's say Car B moves 'x' distance in time t. So, for Car A to cross Car B, it has to move by a distance of 'x+10' in the same time interval t (since Car A has to move an extra distance of the sum of its own length and that of Car B, so as to cross Car B). So, the equation that stands is x/10 = (x+10)/15. Thus from this equation, x is 20. So, road distance covered by Car A is x+10= (20+10)= 30m.
    WHAT MY BOOK SAYS: The answer given in my book is 35m. But i don't know how!
    WHAT I WANT TO KNOW: I just wanna know the flaw in my calculation. To me, my method seems allright.But why is the answer not matching?
    THE PROBLEMS (Contd.): (2) "A police jeep is chasing with 45 km/h.A thief in another jeep is moving with 155km/h. Police fires a bullet with muzzle velocity 180m/s. Find the velocity with which the bullet strikes the jeep of the thief."
    How is it possible to find the final velocity without knowing the distance travelled? I mean the problem doesn't give the initial distance between the two jeeps. ***ERROR PROCESSING*** BRAIN CANNOT SUGGEST ANYTHING***INSUFFICIENT DATA FOR MY BRAIN.
    WHAT MY BOOK SAYS: The solution of this problem is given in my book. It says: "V= v(police) + v(bullet) - v(thief). Putting the values and changing km/h to m/s, the answer given is 150m/s.
    WHAT I WANT TO KNOW: How do we get the above formula(the one given in my book)?
    CONCLUSION: Hey, I'm really confused. Any help will be appreciated. Thanks a lot.
    Last edited: Oct 16, 2006
  2. jcsd
  3. 1) Car A travels at speed of 5m/s wrt car B. Car A has to travel 10m wrt to car B to pass the car B, and it takes 2s to do that. In a meanwhile during that 2seconds, car B travelled 2*10 = 20m.

    So adding 20m + 10m = 30m. That's what you got. Are you sure the textbook answer is right? Can anyone else give me & app any ideas if we are doing something wrong?

    2) The thief car is travelling 155m/s. The police car is travelling 45m/s, so the thief car is travelling 105m/s wrt to the police car. The police fires bullet at the speed of 180m/s, but it only travels at the speed of 180-105= 75m/s wrt the thief car.

    I hope that helped.
  4. I Dont Get The Second Answer.

    I think the first one is ok, i.e. maybe the answer given in the book is wrong. But i dont get your second answer. and its 155 km/h and not m/s. its a jeep, not a supersonic jet. only the velocity of bullet is 180m/s.
  5. haha, true enough... we probably won't live long enough to see a jeep capable of travelling at 155m/s.

    Anyway, just do a conversion of speed of bullet into km/h, then do the same thing. Basically the jeep is faster than the police car. If you think the police car is stationary (i.e. v = 0km/h), then the jeep is travelling at 105km/h. The bullet speed is measured wrt to the police car, so the actual speed of the bullet (i.e. in the lab frame of reference) it is 180m/s + 45km/h (whatever that works out to).

    Draw a little diagram if you haven't done so yet. it always helps me when doing questions like this.
  6. I agree, the book answer is wrong.

    For the second problem, kcirick's idea is right, he just made a mistake with the units. The book is correct here.

    The velocity of the bullet is as it comes out of the muzzle of the gun. If the guns muzzle is already traveling at 45 km/h, then the actual velocity of the bullet (with respect to the ground), will be 180 m/s + 45 km/h.

    If you can throw a ball standing still at 10 m/s, and you can run at 10 m/s, then you can throw a ball at 20 m/s by throwing it while you are running. It's the same idea.

    But the thief's car is also moving, away from the bullet. So the actual velocity of the bullet, relative to the moving thief's car, is reduced by the velocity of the moving target.

    So if you can catch a ball at 10 m/s, but you can't catch it at 20 m/s, just run at 10 m/s in the direction of the ball. Then it's velocity, relative to you, will be 10 m/s and you can catch it.

    Hope this was helpfuly.

  7. Thanks...

    THANK YOU: Hey, thanks a lot for helping. Now I got the second one. But are you sure the book's answer for the first one is wrong? I am quite sure it is wrong. But was just being surer. Anyway, thanks again.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook