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3 physics problems

  1. Sep 14, 2006 #1
    Hello guys, I'm taking my first physics class in College and wanted to cross reference my homework. So here goes :)

    here's question 1:

    A particle moves according to the equatio x=10t^2 where x is in meters and t is in seconds.

    a) find the average velocity for the time interval from 2.00s to 3.00s.

    b) Find the averege velocity for the time interval from 2.00s to 2.10s.

    My answers:

    a) [10(2.00)^2]-[10(3.00)^2)] = -50 (displacement)
    than to find velocity which = displacement/time = -50m/1s = -50m/s

    b) I pretty much did it the same way as answer way, only I plugged in 2.10s and 2.00 into the equation to give me 4.1m/s

    Question 2

    create a graph using these figures to make a position to time graph:

    t(s) 0......1.0.......2.0.......3.0.......4.0........5.0
    x(m) 0..... 2.3......9.2......20.7.....36.8.......57.5

    a) construct a smooth graph

    b) by constructing tangents to the (x)t curve, Find the Instantanious Velocity of the car at Several instants.

    c) Plot the Instantaneous velocity versus time and, from this, determine the average accelleration of the car

    d) What was the initial velocity of the car?

    my answers:

    a) I constructed a graph which slopes upwards (and which looks quite sexy by the way :lol)

    b)Now, the way I solved this was by using displacement/time. Don't know if this is right approach. so for t=1s, I got 2.3 s/m. For t=2s, I got 4.6 s/m, and for t=3. I got 6.9 m/s.

    c) now for question c, I have absolutely no idea how to get a velocity graph from the position time graph I used. Can someone tell me how? Because If I know how to get a velocity graph, I'd know how to find the average velocity.

    d) I assumed that this was a trick question, as I assume that the initial velocity is 0.

    Question 3, last and final question:

    A hare and a tortoise compete in a race over a course 1.00 km long. The tortoise crawls straight and steady at it's maximum speed of 0.200 m/s towards the finish line. The hare runsits maximum speed of 8.00m/s towards the goal for 0.800 km and than stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race, which the tortoisewins a photo finish? Assume that when moving, both animals move steadily at their respective maximum speeds.

    My answer:

    I constructed my own graph for this that I can't really put up on here, unfortunately, But I'll try to walk through what I did.

    I first converted 1km to 1000m. I took the hair and too 800m/8m/s which gives me 100 sec. I than took 1000m/0.200m/s which gives me 5000 sec, which tells me how long it'll take for the turtle to reach the finish line. Now from 800m to 1000m means it'll take 200m for the hare to get to the finish line, in 25 sec from the 800m mark (or 0.800 km) it'll take 1000 sec for the turtle to make it to the finish line.

    now, since the turtle only has 25 sec to make it to the finish line, I multiplied 0.200 m/s by 25s to give me 5m. So the turtle can be 5 m from the finish line before the hare can continue to run to win the race.

    right? lol

    If someone could cross check these questions for me it would be greatly appreciated, as I hope this site will help me with my homework adventures in physics :)

    thanks guys!!!

    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 16, 2006 #2

    For 1a), I think you have [10(2.00)^2]-[10(3.00)^2)] the wrong way round, it should be final minus initial, giving 50m/s. For 1b), I think you missed dividing it by the time. It has a different time interval to the first part, 2.10-2.00=0.10.

    I'm not sure about your answer for 2b). The velocity is given by the gradient of the tangent to the graph at that point, so you need to draw a triangle with the tangent as the hypotenuse, and find the gradient of the hypotenuse using delta(x)/delta(t). For 2c), you need to take your answers to part b), and plot them against time, v up, t across. Then do a gradient triangle to find the acceleration, delta(v)/delta(t).

    Q3) looks fine to me.
  4. Sep 16, 2006 #3


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    Complementing Tomsk's answer:
    For item d) in question 2, the initial velocity is the tangent of your curve at t = 0.
  5. Sep 16, 2006 #4

    cool! Thanks guys, I'll look it over and take your suggestions! Thanks!!!
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