Basic Kinematics Question

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Lunasly

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Hello everyone,

I have a basic kinematics question that keeps bugging me. I often see question that go along the lines of: A constant accelerating particle travels for 3m. If its initial velocity is 10 m/s and it travels 75m, what is its acceleration?

Now solving for the acceleration isn't a problem (I got 10 m/s/s). However, its the concept behind how far the particle has traveled over the 3s that gets me. To my understanding, the particle has an initial velocity of 10 m/s and every second after that it accelerates at a rate of 10 m/s. Since the particle travelled for a total time of 3 seconds, then that means at t=0 we are at 0m, at t=1 we are at 20m (because we are moving at 10 m/s + accelerating at 10m/s/s), at t=2 we are at 50m (because our velocity has increased to 30 m/s). at t=3 we are at 90m because we have accelerated to a velocity of 40 m/s.

Now what I don't get is, how can I say we travelled 90m if the question says we actually only moved 75m?

Thanks,
Lunasly.
 
nu-uh. You are mistaken about one thing. If acceleration is equal to 10, and initial velocity is zero, how much distance is covered?

d = 1/2 at^2 = 1/2 (10)(1) = 5 m

Your mistake is thinking it is 10m. So give it another try.

Average velocity is what you have to consider. (Vf+Vi)/2 = Vavg
 
Hmm, I am still confused. My whole thinking is that if I begin at a velocity of 0 and accelerate at 10m/s every second, then after 1 second I have moved 10m as my velocity is at 10 m/s. The second after that I am now moving at 20 m/s so I must have had to move an additional 20m within that second.

Where am I going wrong? Why does the formula you proposed say otherwise?
 
Hmm, I am still confused. My whole thinking is that if I begin at a velocity of 0 and accelerate at 10m/s every second, then after 1 second I have moved 10m as my velocity is at 10 m/s. The second after that I am now moving at 20 m/s so I must have had to move an additional 20m within that second.

Where am I going wrong? Why does the formula you proposed say otherwise?

That's exactly it. If you begin at 0, and accelerate at 10 m/s^2 every second, then after 1 second you have gained 10 m/s. You have gained 10 m/s VELOCITY, not distance!

So how much distance have you gone in that 1 second, accelerating at 10m/s^2?
Vi = 0
a = 10
t = 1
d = ?

try to use kinematic equations to find the answer to this question. the answer should help you understand why what you wrote is incorrect.
 
Right I understand that my velocity has increased by 10 m/s, but now if I am moving 10 metres every second, doesn't that mean that after 1 second passes, that I have moved 10m?
 
nooo solve the question i posed o_o

it was meant to be didactic

like, with equations and numbers and stuff.
 
nooo solve the question i posed o_o

it was meant to be didactic

like, with equations and numbers and stuff.

Oh sorry I understand that you don't come up with the same answer, but my question is why. Where is my conceptual understanding of the topic wrong?
 
Oh sorry I understand that you don't come up with the same answer, but my question is why. Where is my conceptual understanding of the topic wrong?

The reason it's wrong is because you are thinking in one second blocks. Velocity is gained continuously.

What you're saying would be true if the left graph were true.

Repeating for contrast:

If you are traveling at 10 m/s, with 0 acceleration, then you travel 10 meters in 1 second.

If you are traveling at 0 m/s, with 10 acceleration, then you travel 5 meters in 1 second.

When starting from 0, you are not traveling 10 m/s throughout that entire one second. You start from 0, you go faster and faster until you hit 10.

When you go 10 m/s you start at 10 m/s and you end with 10 m/s. You have a big headstart because you are starting at 10, not at 0.
 
Ah right that makes sense. Velocity is gained continuously.

Thank you so much. I appreciate the help!
 
Very good point. Similar frustrations a few centuries ago lead to the invention of calculus. 😉
 
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