A ball travelling at 6.0 m/s stops over a distance of 1 mm. What is its average deceleration?

Question

Alternatively, what is the average acceleration experienced by the ball to stop it? I think the following formula might need to be rearranged to solve for a.

v^2= v0^2+2ax

v0 = starting velocity
v = ending velocity
a = acceleration
x = distance

I am really bad at rearranging formulas, I managed to get a for acceleration on the left side of the equation as the denominator of v^2, but then I don't know how to get rid of v^2 so a for acceleration is by itself and ready to be solved. Any help would be appreciated.

Answer 1

Yes that is the equation to use. But bear with me a minute as I show that.

v = v0 + at and t = (v - v0)a
x = v0t + (1/2)at^2
x = v0(v - v0)/a + (1/2)a(v- v0)^2/a^2
x = vv0/a - v0^2/a + (1/2) (v^2 - 2vv0 + v0^2)/a
2ax = 2vv0 - 2 v0^2 + v^2 - 2vv0 + v0^2
2ax = v^2 - v0^2
v^2 = v0^2 + 2ax

v^2 - v0^2 = 2ax ..... subtract v0^2 from both sides
(v^2 - v0^2)/2x = a .. divided each side by 2x

v = 0
v0 = 6 m/s
x = 1mm = .001 m

-v0^2/2x = a
-(6)^2/(2*.001) = a
a = -18000 m/s^2

check:
v = v0 + at or v0 = -at and t =-v0/a
t = 6/18000 = 3.333e-4 seconds

x = v0t + (1/2)a t^2
.001 = 6(3.333e-4) + (1/2)(-18000)(3.333e-4)^2
.001 = .002 - .001 = .001

Answer 2

v² = u² + 2ax

u = starting velocity
v = ending velocity
a = acceleration
x = distance

v² = u² + 2ax
v² - u² = + 2ax
a = (v² - u²) / 2x
a = (6² - 0²) / 2(1 x 10^-3)
a = (36 - 0) / (2 x 10^-3)
a = 36 / 0∙002
a = 18,000 m/s (deceleration).

Answer 3

v^2= v0^2+2ax
V²-v0² = 2ax
(X²-V0²)/2x = a
And you take it from there ☺

P.S. You really need to get hot on the Algebra, Pal. It will get a lot deeper than this in a -very- short while ☺

Doug