1. ## displacement formula issue

Hi;
I know that the distance travelled is speed x time.

I've come across another formula for displacement s = ((u + v)t) / 2
where u= initial velocity and v = final velocity and s = displacement.

ex a football is kicked with an initial velocity on the horizontal of 17.63m/s and travels for 3 seconds. Find the distance travelled by the ball

speed x time = 17.63 x 3 = 52.89m.

s = ((u + v)t)/2 = s = ((17.63 + 0)3)/2

s = 13.2225m.

where am I going wrong?

Thanks.

2. ## Re: displacement formula issue

Originally Posted by anthonye
Hi;
I know that the distance travelled is speed x time.

I've come across another formula for displacement s = ((u + v)t) / 2
where u= initial velocity and v = final velocity and s = displacement.

ex a football is kicked with an initial velocity on the horizontal of 17.63m/s and travels for 3 seconds. Find the distance travelled by the ball

speed x time = 17.63 x 3 = 52.89m.

s = ((u + v)t)/2 = s = ((17.63 + 0)3)/2

s = 13.2225m.

where am I going wrong?

Thanks.
Who said the football had no velocity after it dropped for 3 s?

What you are looking for is not the distance the ball traveled. (Unless your instructor is a putz anyway.) What you are looking for is the horizontal distance the football has traveled for 3 s. So what horizontal component of the velocity does the football have a t = 0 s? At t = 3 s?

Hint: What is the horizontal component of the acceleration?

-Dan

3. ## Re: displacement formula issue

at t=3 velocity is zero the ball has stopped at t=0 the initial velocity is 17.63.

Thanks

4. ## Re: displacement formula issue

As long acceleration is constant, a, the speed is a linear function of time, $at+ v_0$. For any linear function the average value between two times is just (a+ b)/2.

Specifically, an object with constant acceleration a, velocity, after time t, is $at+ v_0$ and will have gone distance $(a/2)t^2+ v_0t$. Between times $t_0$ and $t_1$, it will have gone distance $at_1^2+ v_0t_1- at_0^2+ v_0t_0= a(t_1^2- t_0^2)+ v_0(t_1- t_0)= a(t_1- t_0)(t_1+ t_0)+ v_0(t_1- t_0)= (t_1- t_0)(a(t_1+ t_0)+ v_0)$. That is the same as an average velocity of $a(t_1+ t_0)+ v_0= \frac{(at_1+ v_0)+ (at_0+ v_0)}{2}$, the average of the beginning and ending speeds.

5. ## Re: displacement formula issue

There is something I'm not getting and I don't know what.
s= ((u + v)t)/2

lets say I roll a ball along the floor at 20m/s for 10 seconds at which it comes to a stop.
s = ((20 + 0)20)/2
s = 400/2 200m is this correct?

Thanks.

6. ## Re: displacement formula issue

Originally Posted by anthonye
There is something I'm not getting and I don't know what.
s= ((u + v)t)/2

lets say I roll a ball along the floor at 20m/s for 10 seconds at which it comes to a stop.
s = ((20 + 0)20)/2
s = 400/2 200m is this correct?

Thanks.
How did you get $t=20$?

7. ## Re: displacement formula issue

oops that should be t=10.

which would make it 100m is that correct?

8. ## Re: displacement formula issue

Projectile Motion Formulas

9. ## Re: displacement formula issue

Originally Posted by anthonye
oops that should be t=10.

which would make it 100m is that correct?
Yes, that is correct, if t=10. You should attentive :-)

10. ## Re: displacement formula issue

if that's correct why do most posts online use the formula were they cancel out the acceleration term?

11. ## Re: displacement formula issue

Originally Posted by anthonye
if that's correct why do most posts online use the formula were they cancel out the acceleration term?
It only works with constant acceleration. If acceleration is not constant, that formula will not work.

Example: Acceleration = $12-2t$. So, as $t$ changes, so does acceleration. The formula becomes $s = 20t+6t^2-\dfrac{1}{3}t^3$. At $t=0$, $v=20$. At $t=10, v=0$. By the formula you were using, the displacement would still be 100m. But, if you plug 10 into the formula, you find that $s(10) = 200+600-\dfrac{1000}{3} = \dfrac{1400}{3} \approx 467$. That formula for displacement based on average speed is a much more restrictive method of determining displacement. It only works for constant acceleration.

12. ## Re: displacement formula issue

finding this hard to understand.

13. ## Re: displacement formula issue

Originally Posted by anthonye
finding this hard to understand.
If acceleration is constant, you can use the average speed formula you are using. If acceleration is not constant, you cannot.

Thank you.