Motion of an Object: Displacement & Average Velocity

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Stewkatt
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Homework Statement
I drew a time line and tried but i think I missed something can someone try this question and show the steps they took? Thank you!
Relevant Equations
Vav=(vf-vi)/2 or delta d/ delta t
a=(vf-vi)/t
An object starts from rest and accelerates at 3.0 m/s2 for 4.0 s. Its velocity remains constant for 7.0 s, and it finally comes to rest with uniform deceleration after another 5.0 s. Find the following:
a. the displacement for each stage of the motion
b. the average velocity over the whole time
interval.
 
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Let us break the question into bite size pieces. First:
An object starts from rest and accelerates at 3.0 m/s2 for 4.0 s.
What is its position at t=4.0s ?
What is its velocity at t= 4.0s ?
 
Hey, I figured it out. Here is my work.
image.jpg
 
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That's looks correct. Note that you could also have drawn a velocity against time graph and calculated the displacements as the area under each section of the graph.
 
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PeroK said:
That's looks correct. Note that you could also have drawn a velocity against time graph and calculated the displacements as the area under each section of the graph.
Hey, my teacher prefers I solve things with algebra with a timeline, but I’ll keep that in mind
 
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Stewkatt said:
Hey, I figured it out. Here is my work. View attachment 296860
You can make your working a bit shorter.

Stage 2 is at constant velocity (12.0m/s) so using ½(12m/s+12m/s)*7.0s seems a bit excessive.
d₂ = 12.0m/s*7.0s should be acceptable.

For Stage 3. there’s no need to find and use acceleration; just multiply average velocity by time:
##d = \frac {(v_i+v_f)} {2}t##.
 
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The quickest way is to note that the maximum speed (during the middle phase) is ##12m/s## and the average speed during both the acceleration and deceleration phases is half of this. So, the displacements are: ##d_1 = 4s \times 6m/s = 24m##, ##d_2 = 7s \times 12m/s = 84m##, ##d_3 = 5s \times 6m/s = 30m##.
 
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Thank you for all your help everyone :)
 
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