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Problem on distance, time and velocity when acceleration is the only given.

  1. Dec 11, 2008 #1
    Two cars, car A and car B. Car A and car B initially at rest, both starts at the same time. Car A is initially at some distance away behind car B.

    A.) How long does it take for car A to overtake car B?
    B.) How far was car A behind car B initially?
    C.) What is the speed of each car just before overtaking?

    Given:
    acar A= 3.5 m/s2
    acar B= 2.2 m/s2
    tcar A=tcar B because they started on the same time.
    vi of car A=vi of car B=0 because they are initially at rest.

    Please someone help me with this?
     
    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 11, 2008 #2
    Please check my solutions regarding another problem.

    Boy A is on the roof, 46m above the ground. Boy B is sneaking slowly alongside the tower at a constant speed of 1.2m/s. How far should boy B from the tower when boy A releases the stone from the roof to make a sure hit?

    Given:
    Boy A on tower
    vi=0
    yi=0
    yf=46m
    g=-9.8m/s2

    Boy B alongside
    xi=0
    v=1.2m/s
    a=0 because boy B has a constant speed.

    Solutions:
    To solve for time t, I used the formula yf=vit-1/2gt2. y=-1/2gt2
    -2y=gt2
    -2y/g=t2
    sqrt(t2)=sqrt(-2y/g)
    t=sqrt[-2(46m)/(-9.8m/s2)]
    t=3.06399437s

    Next question, How far should boy B from the tower when boy A releases the stone from the roof to make a sure hit? I used the formula xf=vit+1/2at2.
    xf=(1.2m/s)(3.06399437s)
    xf=3.676733244m

    Do I get it right? Please make some corrections if I do it the wrong way. Thank you again.
     
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