what am i doing wrong
1
smart VLR users come pt3
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DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
s stands for distance
part1: 0 - 10s
s: 1 x 100 = 100m
100m = s
v: 100 = v^2 / 4
v: 400 = v^2
20m/s = v
part2: 10 - 15s
s: 0 x 25 + 20 x 5
s: 20 x 5
100m = s
xxvelocity stays same as no acceleration (v = 20m/s)xx
part3: 15 - 20s
s: -1.5 x 25 + 20 x 5
s: -37.5 + 100
62.5m = s
part1 + part2 + part3 = 100m + 100m + 62.5m + = 262.5m
−11.6 cm/s^2 is what i got but idk if i did the math correct. the idea should be right though.
i used the one of the uniform acceleration equations, displacement = velocity initial time + (1/2) acceleration * time^2 (d = vit + (1/2)at^2). Everything is given to you- the displacement in the x-direction is xf-xi. you have the time interval and initial velocity, so just solve for acceleration.
It's pretty simple with the four main kinematic constant acceleration equations, you just have to get the idea down.
iirc vi*t calculates the intial displacement at the initial velocity over the time interval. But because the object is accelerating, you are gaining more displacement at the increasing speeds over the time interval. Think of it as adding the original displacement of the object if it were traveling at the initial speed (11 cm/s) for 2.45s. But since the object gains velocity from that initial velocity at an acceleration, 1/2at^2 calculates such displacement.
imagine the object traveled at 11cm/s for 2.45s and displaced 26.95 cm. Now imagine we travel at the same speed and displacement right next to the object, treating speed relative. We are both currently traveling at a constant velocity. the object in this scenario gains an imaginary little hand that applies a force and accelerates it. We would begin to see the acceleration because the object gains velocity (in this example we would see the object slow down relative to us because it accelerates in the negative x-direction). That negative displacement from the acceleration is the 1/2at^2 we add to the original displacement at 11cm/s for 2.45s.
You're correct I just think you didn't step it out in a very clear way.
You should start most problems like this by writing out the information you have in a clear way so you can figure out what formula you want to use.
In this case we have: t = 0s, where vi = 11cm/s, and d = 2.91cm
then we also have : t = 2.45s , vf = ? and d = -5 cm (note we don't need the final velocity)
It also says in the question that acceleration is uniform (so a constant)
You probably have a formula list somewhere and from that you can figure out the one you need is the one below.
Δd = vi x t + 0.5 x a x t^2 (delta d = total distance moved, vi = initial velocity, a = acceleration, t = time)
a is what we are trying to solve for and delta d is the total change in position so Δd =(-5) - 2.91 =-7.91
plugging everything into the formula we get the following.
-7.91 = 11 x 2.45 + 0.5 x a x 2.45^2
(-7.91-11 x 2.45)/(0.5x2.45^2)=a
a=-11.615cm/s^2
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