Sorry, guys, all wrong!!! Let an old fart do it right......

Start with the distance-time-acceleration formula:

d = 0.5 * a * t^2 ...

EQ(1)

d = distance = 100m

a = acceleration = ?

t = time to decelarate from 40 m/s to 0 = ? ( t^2 = t "squared")

Next, the definition of acceleration is rate of change of speed ..... at constant acceleration, it is simply the change in speed divided by the time taken for that change .....

a = (Vf - V0) / t ...

EQ(2)

Vf = final velovity = 0 m/s

V0 = initial velocity = 40 m/s (by the way, velocity is m/s, not ms!)

Note that we have velocities and distance, want to solve for acceleration. So let's eliminate time from equation (1) by substituting equation (2), thus ....

Rewriting

EQ(2):

t = (Vf - V0) / a ...

EQ(3)

Substitute for t in

EQ(1):

d = 0.5 * a * { (Vf V0) / a }^2

Collect terms, simplify:

d = 0.5 * (Vf - V0)^2 / a or,

d = (Vf - V0)^2 / (2 * a)

Now, rearrange to solve for a:

a = (Vf V0)^2 / (2 * d)

Finally, plug in the numbers:

a = (0 - 40)^2 / (2 * 100)

= 1600 / 200

= 8 m/s^2 (metres per second per second)

Now, to check:

Substitute into equation (3):

t = (0 - 40) / 8 = 5 seconds

Substitute into equation (1), using t = 5 s:

t = 0.5 * 8 * (5)^2

= 0.5 * 8 * 25

= 100 metres

It checks out!!

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Now, Fritzi was almost there .... he solved for time correctly (5 sec) but inadvertantly called it acceleration ... see above ... he gave T = 5 m/s (sq) should have been 5 sec.

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Here's a trick: you can easily determine if an equation is NOT correct if the units don't work out. For example, SunAir gave v=u + 2as (v, u are velocity, a is acceleration, s is distance) - here's how to check it out .....

Substitute units for variables:

v = u + 2as

m/s = m/s + (2) m/s/s * m {note the constant 2 has no units - just ignore it}

Note the last term, when you simplify, becomes m*m /s/s or (m/s)^2

That is velocity squared, and is inconsistent with the other terms which are velocity.

This methodology is a part of "dimensional analysis", which is used in some sciences to construct equations where physical models are not well understood.

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You second problem can also use equation (1), but you have to rewrite it as a function of time:

t^2 = d / (0.5 * a) but a = accel of gravity, or g = 10 m/s^2

t^2 = d / (0.5 * g)

t = sqrt { d / (0.5 * g) }

Plug in numbers ......

t = sqrt { 3 / (5) } = sqrt { 0.6 } = 0.775 seconds

Fritzi is correct!!!

I hope you can follow this. Please don't just copy it, make sure you understand it. I'll be glad to answer any questions. I'm working on my taxes today, so I'll be on my computer off and on all day.

Good luck,

Pete