Silly Kinematics Question

[unleash500]

Why does
x= 1/2(vo + v) t?

I understand that it is an "average" , but it is conflicting with my normal every day intuition. Let's say I drive 10 secs at a velocity of 10 m/s, and that my initial velocity was 0 m/s. My intuition tells me that I have traveled 100 m, but based on the equation above I would have traveled 50 m right?

For some reason I feel like I just asked a very stupid question lol.

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[MLT2MT2DO]

Why does
x= 1/2(vo + v) t?

I understand that it is an "average" , but it is conflicting with my normal every day intuition. Let's say I drive 10 secs at a velocity of 10 m/s, and that my initial velocity was 0 m/s. My intuition tells me that I have traveled 100 m, but based on the equation above I would have traveled 50 m right?

For some reason I feel like I just asked a very stupid question lol.

You are forgetting to add the time it took to go from 0-10 m/s. What you did was go from 0-10m/s in 10 seconds, in which case you would have only traveled 50m.

The way your using your intuition is not the formula. You are using your intuition as you are starting out at 10 m/s and continuing to go 10 m/s for 10 secs. If you plug 10 into both your vo and v you now have the 100 meters that you're talking about

 

[kentavr]

This formula is right only when acceleration is constant. That means that you increase your speed linearly or V(t) = V(0) + A*t. Where A-is a constant acceleration. If you assume that the acceleration is not constant and has more complicated form, then this formula is not correct anymore.

 

[surag]

Yah but 10 * 10 only works if u travelled at 10ms for 10 seconds. But u didn't. U started at 0 so obv u were changing speeds till u reached 10. In essence u travelled at 10ms for only 1 second!

 

[boggvir]

Why does
x= 1/2(vo + v) t?

I understand that it is an "average" , but it is conflicting with my normal every day intuition. Let's say I drive 10 secs at a velocity of 10 m/s, and that my initial velocity was 0 m/s. My intuition tells me that I have traveled 100 m, but based on the equation above I would have traveled 50 m right?

For some reason I feel like I just asked a very stupid question lol.

You have to do calculus in cases where the acceleration is not constant. If you have non-constant acceleration, you have to deal with the physics term known as jerk (which is the rate of change of acceleration). In mathematical terms, jerk is the derivative of acceleration with respect to time:

j = da/dt

The units being m/s^3 (instead of m/s for velocity, m/s^2 for acceleration).

However, for the situation you described, it is simpler to model it in terms of two phases: phase one where your car was going from 0-10 under constant acceleration, and second where it was travelling with constant speed. You'll get the right answer doing it that way instead of getting into calculus.

For the MCAT, if you have to do calculus, you're doing it wrong .