Starting from rest, the distance covered in t seconds by an object accelerating at a constant a m/s/s is (a*t^2)/2
An object travelling at a constant b m/s covers a distance b*t in t seconds,
These are both examples of the general equasion:
d = d(zero) + v(zero)t + (1/2)at^2.
What we know is that at the point where the automobile will pass the motorcycle, the two have travelled the same distance from the stoplight.
Thus, d(auto) = d(motorycle).
And the "show all work" form becomes:
d(auto-zero) + v(auto-zero)*t + (1/2)a(auto)*t^2 = d = d(mc-zero) + v(mc-zero)*t + (1/2)a(mc)*t^2
Known:
d(auto-zero) = zero, because we are the ones establishing our coordinates.
d(mc-zero) = zero (same reason)
v(auto-zero) = zero (as per instructions)
v(mc-zero) = 15 m/sec (as per instructions)
a(auto) = 1.5 m/sec^2 (as per instructions)
a(mc) = 0 m/sec^2 (as per instructions)
t = unknown
Yup, which then in the "showing all work" mode, you then divide both sides by t:
(1.5/2)t^2/t = 15t/t
...and then multiply by 2 (because its more obvious this way):
1.5t = 15*2
...and divide by 1.5:
t = (30)/(1.5)
Same answer I got.
FWIW, the first general formula listed above, namely:
d = d(zero) + v(zero)t + (1/2)at^2.
...is the
second derivative of distance with respect to time. If you've had Calculus already, this is an "oh yeah" and you can see how to derive it. Otherwise, you'll just have to memorize it.
The reason I mention this is because of this physics formula which should also look familiar:
v = v(zero) + at
This is the
first derivative of distance with respect to time, and it will also probably show up on a Physics exam problem.
In general, the 'trick' to solving these problems is to keep yourself organized by laying out what data you need to solve whichever formula, then how the facts that you know will fit together to allow you to solve simultaneous equasions.
For example, Part II of the above question could be to ask:
"...and what was the velocity of the automobile at the time that it catches up with the motorcycle?"
The answer here is pretty simple: its "v = v(zero) + at" when t = 20sec.
Similarly, Part III could have been:
"...and what was the distance at which the automobile catches up with the motorcycle?"
Answer here is to plug t=20 into the d = d0 + V0t + 1/2at^2 formula.
Be aware that really clever tests will ask you just this second or third part and not give you the hint as to what you need to solve for first (here, we used time). Just lay out the facts in an organized fashion and you'll figure out where there's gaps that either have to be solved for, or for which you need to figure out how you can mathematically eliminate them.
-hh
