In the particle under constant acceleration model, we identify the variables and parameters vxi, vxf , ax, t, and xf − xi. Of the equations in the table below, the first does not involve xf − xi, the second does not contain ax, the third omits vxf , and the last leaves out t. So, to complete the set, there should be an equation not involving vxi.
(a) Derive it from the others. (Use any variable or symbol stated above as necessary.)
xf − xi =
(b) Use the equation in part (a) to solve the following problem in one step. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.53 m/s2 for 4.01 s, making straight skid marks 60.1 m long, all the way to the tree. With what speed does the car then strike the tree? (in m/s)
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Verified answer
Who asks you this garbage? A table with four constant-acceleration equations? Complete the set? What an enormous waste of time and effort. And further, it obscures the fact that physics is about ideas, and turns it into some farcical exercise in memorizing equations.
There are two constant acceleration equations that are worth remembering. You can always decide not to remember any, and derive these two using calculus, but assuming you don't want to start every constant-acceleration problem with calculus, you should know
xf = xi + vxi t + ½ ax t²
vxf = vxi + ax t
all the rest of "the set" can be derived using algebra.
To eliminate vxi, I'd solve the second equation for it
vxi = vxf - ax t
and substitute it into the first one:
xf = xi + (vxf - ax t) t + ½ ax t²
I suppose the creators of idiotic tables would prefer this to be simplified
xf = xi + vxf t - ½ ax t²
So that's part "a". Then in part "b" we're supposed to use this to solve a problem "in one step", as if all we did in part "a" somehow didn't count as "steps", and as if solving a problem in one step were a laudable goal, instead of a sign that whoever's asking the question has missed the point of studying physics and problem solving.
Choose a coordinate system. The sign of the given acceleration tells us that the positive direction must be opposite the original (and presumably the final) velocity of the car. Let's put the the origin at the point where the brakes are applied. With that coordinate system:
xf = +60.1 m
xi = 0 m
vxf = ?
ax = -5.53 m/s²
t = 4.01 s
Solve
xf = xi + vxf t - ½ ax t²
for vxf
vxf = [ xf - xi + ½ ax t² ] / t
vxf = [ (+60.1 m) - (0 m) + ½ (-5.53 m/s²) (4.01 s)² ] / (4.01 s)
vxf = +3.9 m/s
which is the answer to part "b". Whoever asked you this question should go away, and not come back until they've given some serious thought to their goals in providing physics instruction. No wonder students hate physics and think it is unreasonably difficult.