# Why is d=vt+1/2a(t)^2

1. Nov 14, 2015

### redhatbeast

Im in high school and new to physics. according to d=vt+1/2a(t)^2, if we have a car that is not moving and after 1 second it reaches the speed of 10 m/s and the acceleration is 10 m/s then after 1 second it traveled 5 meters.
How is it possible that it traveled 5 meters after 1 second?

2. Nov 14, 2015

### Staff: Mentor

Hi redhatbeast, welcome to PF!

You have the correct formula, you have applied it correctly, and you have obtained the correct answer. I am not sure what else you are asking. Did you just want confirmation?

3. Nov 14, 2015

### Staff: Mentor

What is the average speed over the 1 sec?
(a) 0 m/s
(b) 10 m/s
(c) 5 m/s

Chet

4. Nov 14, 2015

### redhatbeast

But what im asking is that if you imagine a car that is not moving and then after 1 second that it starts moving it will reach speed of 10 m/s. So the distance that it traveled after 1 sec should be more than 10 meter. But with the formula you get 5m which doesnt make sense. Can somebody explain to me how is that possible that it travels 5 meter?
Thanks!

Last edited: Nov 14, 2015
5. Nov 14, 2015

### ehild

Why do you think that?

6. Nov 15, 2015

### redhatbeast

If we draw it on a graph and we calculate the distance using pythagorean theorem its √2. So its greater that 1 meter.

7. Nov 15, 2015

### Staff: Mentor

That's not the correct way to find or graph position/distance. Notice that in your graph, if you are initially moving at 1 m/s with 0 acceleration, then your x-position would not be increasing at 1 meter per second, but something else. The Pythagorean theorem simply can't be used for this.

8. Nov 15, 2015

### sophiecentaur

How did you arrive at that answer. I think you have been more intuitive than is appropriate. The sums are very straightforward and give you 5m.
The graph that you have drawn is not the right one. Look at this link to find the way to approach things. (Clue: it's a matter of area)
There is no arm waving answer to your "why" question. The maths is a perfectly valid description of the process.

9. Nov 15, 2015

### Staff: Mentor

I'll try again. This time, please don't disregard my questions. You are driving 30 km/hr for the first hour, and 90 km/hr for the 2nd hour. What is your average speed, and how far have you traveled all together? Do you need to use the Pythagorean theorem to get this answer?

Chet

10. Nov 15, 2015

### Staff: Mentor

Does the Pythagorean theorem make sense? Check your units at each step and see if they are compatible.

11. Nov 15, 2015

### sophiecentaur

He travels 25kg at a speed of 43°C?

12. Nov 15, 2015

### redhatbeast

Thanks everyone,
you want to find the average speed, for example the average speed of 0 and 10 m/s is 5 m/s. Then you just multiply that by time (1s) and you get 5m which is the distance traveled.
Thanks everyone!

13. Nov 15, 2015

### sophiecentaur

Oh nasu - you can be nearly as childish as me! (correction: "I")

14. Nov 15, 2015

### Staff: Mentor

You posting in the right thread, Sophie?

15. Nov 16, 2015

### sophiecentaur

The right thread is always the one I am on!!
Nasu just liked my inane post, above, and I thought it deserved a reply in the appropriate vein.

16. Nov 16, 2015

### nasu

Oh, thank you for explaining. I was just wondering what have I done.

17. Nov 16, 2015

### sophiecentaur

You just appreciated a naff joke. (Probably a bannable offence) v

18. Nov 16, 2015

### nasu

Well, it was not (just) the joke but the way you use the joke to bring up a point related to the OP's graph.

19. Nov 20, 2015

### Chuckstabler

Calculus. The derivative of y with respect to x is the slope of a line tangent to the curve of y at point x, or the rate of change in y at point x of the function y(x). The rate of change of distance is simply velocity. Therefore the derivative of distance (as a function of time) is velocity. The rate of change in velocity is simply acceleration (once again as a function of time), so the derivative of velocity is acceleration. Now, the derivative of y with respect to x of y = x^n is n*x^(n-1). So if we start with x^n and want to find a function whose derivative is x^n. the derivative of x^(n+1) is (n+1)x^n, so the derivative of (x^(n+1))/(n+1) is x^n (since the derivative of some constant * f(x) is just that constant times the derivative of f(x). Knowing this, we can move backwards from the acceleration to velocity. If we have a constant acceleration A, then we know that the derivative of our velocity function = A. We can think of this as A*t^0, since t^0 = 1. So the function whose derivative is A is A*t^(0+1) divided by (0+1), which equals A*T. Now, the derivative of a constant is 0, so we have to add a constant to our velocity function to account for the range of possible functions. So velocity equals A*T + C1. We solve for this constant by setting the velocity at some point to some value, called V0. We usually set it equal to this value when time equals 0 because thats the point that our function "starts". So v0 = A*0 + C1, so v0 = c1 so velocity equals A*T + V0. Now we do the same thing to find displacement as a function of time. A*T^(1+1)/(1+1) + v0*T^(0+1)/(0+1) = 0.5A*T^2 + v0*T + C2 = displacement! solve for the constant again to get c2 = d0 = initial displacement.