Alan

That is correct Chris!

There is also enough information to determine that the initial horizontal velocity is 32.9/2.27 ≈ 14.5 m/s.

Or you could write it as (x + √5)²

Just to confirm Melody's answers:

Q2.1   A.  Velocity is slope of x-t graph including sign.   Slope is most positive at P.  

Q2.2   C.  Speed is magnitude of slope of x-t graph (i.e. ignoring sign).  Slope has largest magnitude at R.

Q2.3   D.  Acceleration is slope of velocity-time graph (including sign).  Just before point S the velocity is large and negative; just after point S it is large and positive; therefore the change through S is from large negative to large positive velocities.  This change is positive and is larger at S than anywhere else.

For the middle question the answer is A.  The net force on the mass is in the positive x-direction and is constant (assuming static and moving friction are the same).  With a constant mass, this means the acceleration is constant.  

The psychologically misleading aspect of the question is that it is difficult to think in terms of acceleration; we are more inclined to see think in terms of velocity, so at first imagine the mass slowing down and then speeding up again.  It does this, of course, but we now have velocity in mind instead of acceleration.  Well, that's my excuse!!

Melody is right on the first and third questions.

You can take it just a little further Anonymous:

2log(5) + 2log(2) = log(5²) + log(2²) = log(25) + log(4) = log(25*4) = log(100) = 2 (assuming log to the base ten)

If you hover the cursor over the log key you will see ln, e^x, lg2 and 2^x pop up.  lg2 is log to the base 2.

Alternatively, you can type log(x,2) where x is the number of which you want to find log to the base 2.

There are three forces acting on it.  The force of gravity (mass*g) is one.  The reaction of the road surface on the boulder is another. The static friction of the road surface is the other - this will be given by μ*mass*g*sin(13°), where μ is the coefficient of static friction between the boulder and the road.  The net force is zero, as the boulder isn't accelerating (unless one of the cars slams into it!).

Iodine-131 decays according to the radioactive decay equation N/N₀ = e^-λx, or N/N₀ = e^-ln(2)x/τ, where N is amount, N₀ is initial amount, x is time, λ is the decay constant and τ is its half-life  (λ = ln(2)/τ).  For Iodine-131 the half-life is approximately 192.5 hours.  

So, if we replace N/N₀ by f(x), then b = 1 and a = e^-λ or a = e^-ln(2)/τ  (your choice!).  Plug the numbers into a calculator to find the numerical value of a.

The equation of a straight line can be written as y = m*x + c where m is the slope and c is the y-intercept.

A perpendicular to this will have slope -1/m.

I'm not sure if the following is what you are looking for:

Your equation is y = (1/3)x + 1

The perpendicular will be in the form y = -3x + c

If it goes through (4, 1) then

1 = -3*4 + c so that c = 13

The equation of the perpendicular line passing through (4, 1) is therefore  y = -3x + 13