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Help with displacementby Joules23
Tags: displacement 
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#1
Sep806, 12:29 PM

P: 43

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 


#2
Sep806, 12:31 PM

P: 147

try using the equation x=1/2*a*t^2 for each aceleration and time and add them together.



#3
Sep806, 01:02 PM

P: 43

i got 39.5, 13.56, 20.4, this cant be right



#4
Sep806, 01:08 PM

Mentor
P: 41,579

Help with displacement



#5
Sep806, 01:15 PM

P: 43

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 


#6
Sep806, 01:18 PM

Mentor
P: 41,579

See if this helps: Basic Equations of 1D Kinematics



#7
Sep806, 01:38 PM

P: 43

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? 


#8
Sep806, 01:46 PM

Mentor
P: 41,579

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] 


#9
Sep806, 02:07 PM

HW Helper
P: 3,220

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.



#10
Sep806, 02:33 PM

P: 43

im LOST
displacement= XX0.. i dont understand how i find X0 or X 


#11
Sep806, 03:13 PM

P: 43

XX0=V0t + .5at^2
XX0=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 


#12
Sep806, 03:50 PM

HW Helper
P: 3,220

Use the diagrams. As mentioned, calculate the array of the vt 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 vt diagram.



#13
Sep806, 04:03 PM

P: 43

ur saying i need to find the resultant vector?



#14
Sep806, 04:20 PM

HW Helper
P: 3,220

No, there are no vectors. Just calculate the area of the vt diagram, and that's your b) result.



#15
Sep806, 05:02 PM

Mentor
P: 41,579

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.) 


#16
Sep806, 10:05 PM

P: 43

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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