This lesson will be a combination of the two previous ones, but will introduce some
new formulas. Just so you know, this lesson shows just about every step in solving this
problem. For those of you who like to see every step to make sure you're not missing
anything, this is for you. For those of you who don't like to see everything worked
out, please bear with me. Now, instead of beginning with the formulas, let's begin with
a problem.
A motorcycle travels the first half of a 100m track with a constant speed of 5m/s.
In the second half of the 100m, it experiences a flat tire and slows down at a rate of
0.2m/s2. How long does it take the cycle to travel the 100m distance?
Ok, seems easy enough. Let's start with the first half of the track. The bike
traveled half of 100m at 5m/s.
100m / 2 = 50m
50m / 5m/s = 10s
t1 = 10s
Now for the second part. The bike travels the last 50m starting at 5m/s and
accelerating at -0.2m/s. We have distance, initial velocity, and acceleration, and we
need to find time. But wait, none of our formulas can do that in one easy step!! Nope,
this one's not a one shot deal, we have to substitute equations.
First, put what we know into the equations.
-0.2m/s2 = (vf-5m/s)/t
50m/t = (5m/s+vf)/2
Now, since we want time, let's substitute for vf.
vf = -0.2m/s2*t + 5m/s
vf = 100m / t - 5m/s
-0.2m/s2 * t + 5m/s = 100m / t - 5m/s
Now we solve for t.
-0.2m/s2 * t + 10m/s = 100m / t
t*(-0.2m/s2 * t + 10m/s = 100m / t)
-0.2m/s2 *t2 + 10m/s*t = 100m
0 = 0.2m/s2*t2 - 10m/s*t + 100m
Oh boy! Quadratic formula!!
(10m/s +- squareroot (100m2/s2 - 4*(0.2m/s2)*(100m))) ÷ 2(0.2m/s2)
(10m/s +- squareroot (100m2/s2 - 80m2/s2)) ÷ 0.4m/s2
(10m/s +- 4.472m/s) ÷ 0.4m/s2
(10m/s + 4.472m/s) ÷ 0.4m/s2 = 36.18s
(10m/s - 4.472m/s) ÷ 0.4m/s2 = 13.82s
Now we have two possibilities of answers. The only way to know which is the real
one is to find some way to cancel one of them out. In this case, let's check the final
velocity of the motorcycle and see what we get.
50m/36.18s = (5m/s+vf)/2
vf = -2.236m/s
50m/13.28s = (5m/s+vf)/2
vf = 2.530m/s
Since it is impossible to have a negative speed, 36.18s is invalid. Therefore,
13.82s is the answer to the second part. That added to the 10s we got from the first
part is 23.82s, which is the correct answer.
Now, here's an easier way to do the second half. Instead of substituting and doing all
that work at the beginning, here are two equations that have already been substituted
using only variables so you can plug in your numbers and go.
2ad = vf2 - vo2
d = vot + 0.5*a*t2
The second of the two equations applies more to our situation, so we'll use it.
Just plug in the values we already have
50m = 5m/s*t + (-0.1m/s2)*t2
Now do the quadratic formula and get the same answers as before
t = 36.18s
t = 13.82s
We already know that 13.82s is the correct answer, so just add 10s to it and we get
23.82s.
Much less work, wasn't it? These two new formulas should be easier and faster for
you to use. Don't forget the other two you learned though, because in some cases they
will be faster than the two new ones.
The counterpart, of course, to horizontal motion is vertical motion, which is what
the next group of lessons will cover.
Horizontal Motion Sample Problems
