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Motion questions show me the steps

  1. Feb 26, 2005 #1
    Hi guys i've got here motion questions and was wondering if someone could show me what i am doing wrong.

    Q 1. A truck on a straight road starts from rest accelerating at 2.00m/s^2 until it reaches a speed of 20.0 m/s. then the truck travels for 20s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s a) how long is the truck in motion
    b) what is the average velocity for the motion described.

    a) i added up the seconds ie 2m*10 + 20*20 + 5 = 35 seconds i got that right

    b) this i got wrong i tell you what i did i broke up the av velocity into 3 sections so 1 section for the acceleration, 1 section for the 20s constant, and 5 seconds for the stop but got them wrong when i added then / 3 can someone just write down the steps and i will attempt to do this question thanks.

    ie i did
    step 1) average velocity = 10 m/s
    v=u + at
    v=0 + 2(10)
    v=20 m/s

    20 + 0 / 2 = 10m/s

    step 2) average velocity= 20 m/s as velocity is constant

    step 3) vf=0 vi=20
    average velocity for section3 is 20/5=4 m/s

    20+10+4 / 3 = 11.3 m/s that is incorrect the answer is 15.7 m/s
    *** so someone please write down the steps and i will attempt to do it ok tahnks a lot every1 ***

    Question 2 /

    An inquisitive physics student and mountain climber climbs a 50m cliff that overhangs a calm pool of water. he throws 2 stones vertically downward, 1s apart, and observed that they cause a single splash. The first stone has an initital speed of 2 m/s a) How long after release of the first stone do the two stones hit the water b) what was the initital velocity of the second stone c) what is the velocity of each stone at the instant the two hit the water

    ok for this question i could not do any of the parts i tried to find the time but was unsucessfull i will show you what i did.

    i got confused with the time when the first stone dropped down so i use the formula x=ut + .5at^2
    i put x as 2+ 2t thus 2 + 2 t = .5(-9.8)t^2
    i solved for t but was incorrect can someone please tell me instructions / formulas to solve for parts a.b.c then i will try to attempt this and post back thank you very much
  2. jcsd
  3. Feb 26, 2005 #2


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    Question 1:

    Find the distance travelled in each of the stages. Add them up, divide by the total time in motion.

    BTW - In step three, you calculated the rate of deceleration, not the average velocity.
    Last edited: Feb 26, 2005
  4. Feb 26, 2005 #3


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    Question 2:

    The pool of water is your best reference point (both stones arrived here at the same time). In other words, the water is 0 position while the top of the cliff is s//b]

    You use your equation for position:

    [tex]s_f=s_i + v_i t + \frac{1}{2} a t^2[/tex]

    s_f is the pool of water, or zero.
    Also keep in mind that both v_i and acceleration are negative.

    However long it took for the first stone to drop, it took one second less for the second, so the time for the second stone is (t-1). You plug this in everywhere you had a t in the first equation. It's (t-1)^2, not t^2 -1.

    Final position for both stones is zero, so the two equations equal each other. Set them equal and solve for t. You have two variables (v_i of the second stone and t), so you'll get an expression for t rather than a number.

    Substitute this expression for t in the position equation for the second stone to solve for the initial velocity of the second stone.

    Now, finally, you can solve for the how long it take the first stone to hit the water.

    Knowing the initial velocity of each stone and their acceleration, you can then figure out the final velocity for each stone.
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