Help with displacement


by Joules23
Tags: displacement
Joules23
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#1
Sep8-06, 12:29 PM
P: 45
A speedboat starts from rest and accelerates at +1.87 m/s2 for 6.50 s. At the end of this time, the boat continues for an additional 6.43 s with an acceleration of +0.656 m/s2. Following this, the boat accelerates at -1.43 m/s2 for 5.34 s. (a) What is the velocity of the boat at t = 18.27 s? (b) Find the total displacement of the boat.

I got the answer to
a)
v=ta
v=6.5*1.87=12.155
v=6.43*.656=4.22
v=5.34*(-1.43)=-7.64
v=8.78m/s @ 18.27sec

b)
i cant seem to figure this out. i am trying displacement =vt, but this is getting me only 160, the answer is 198, but i dont know how to get 198
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dmoravec
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#2
Sep8-06, 12:31 PM
P: 147
try using the equation x=1/2*a*t^2 for each aceleration and time and add them together.
Joules23
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#3
Sep8-06, 01:02 PM
P: 45
i got 39.5, 13.56, -20.4, this cant be right

Doc Al
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#4
Sep8-06, 01:08 PM
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Help with displacement


Quote Quote by Joules23
b)
i cant seem to figure this out. i am trying displacement =vt...
The displacement during each segment will equal the average velocity multiplied by the time. What's the average velocity in each segment?
Joules23
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#5
Sep8-06, 01:15 PM
P: 45
i dont see how i can get the average velocity, when average velocity = displacement/time
thats two unknows

when i use the 3 segment velocities i got in part a, it comes to a total of 66
Doc Al
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#6
Sep8-06, 01:18 PM
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See if this helps: Basic Equations of 1-D Kinematics
Joules23
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#7
Sep8-06, 01:38 PM
P: 45
i just tried using this equation, and its not working out:
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

for my first segment, my [tex]X_0[/tex] will be 0 right? but for my second segment the [tex]X_0[/tex] will be my answer to my first segment, correct?
Doc Al
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#8
Sep8-06, 01:46 PM
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That method should work just fine. You can also use the average velocity in each segment to find the incremental displacement (as discussed earlier).

[tex]v_{ave} = (v_i + v_f)/2[/tex]
radou
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#9
Sep8-06, 02:07 PM
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The easiest way to solve this type of problems if to make a sketch of the acceleration and velocity diagrams. I didn't write the values on the diagrams, because I think you'll figure them out easily. I got that the total distance is equal 198 m, and it is represented by the area of the velocity diagram.
Joules23
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#10
Sep8-06, 02:33 PM
P: 45
im LOST
displacement= X-X0.. i dont understand how i find X0 or X
Joules23
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#11
Sep8-06, 03:13 PM
P: 45
X-X0=V0t + .5at^2

X-X0=12.2(6.5) + .5(1.87)(6.5)^2 = 118

Am i doing this right? Please help, ive been trying to figure this problem out for 3 hours now
radou
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#12
Sep8-06, 03:50 PM
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Use the diagrams. As mentioned, calculate the array of the v-t diagram, where the speed at t1 is a1*t1, and so on. After calculating the speeds, you can calculate the area of the diagram. Start with the first triangle. It's area is t1*v1*0.5. The next area is equal t2*v1*+t2*v2*0.5, and so on. The total displacement equals the area of the v-t diagram.
Joules23
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#13
Sep8-06, 04:03 PM
P: 45
ur saying i need to find the resultant vector?
radou
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#14
Sep8-06, 04:20 PM
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No, there are no vectors. Just calculate the area of the v-t diagram, and that's your b) result.
Doc Al
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#15
Sep8-06, 05:02 PM
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Quote Quote by Joules23
X-X0=V0t + .5at^2

X-X0=12.2(6.5) + .5(1.87)(6.5)^2 = 118

Am i doing this right? Please help, ive been trying to figure this problem out for 3 hours now
Here you are calculating the displacement in the first segment. But you are using the wrong value for V0. V0 is the initial velocity, which for this segment is 0 (since it starts from rest). (12.2 is the final velocity.)

To use this method, which is perfectly fine, you'll need the initial velocities for each segment of the motion. (Use what you've already figured out in post #1.)
Joules23
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#16
Sep8-06, 10:05 PM
P: 45
Thanks all Got it!, substitute that for 0, then for the next segment use 12.2, and for the third use 16.42.... thanks again (along with the other substitutions, a, t,v)


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