PH 221
Homework 2 Solution
Winter 2017
Chapter 2
Two cars move in opposite directions toward each other on a level, straight, one-lane
road. When both drivers begin to brake, their front bumpers are 275 meters apart. The
smaller car is initially moving at 25.0 m/s and the larger is initially moving at 30.0 m/s.
The magnitude of the acceleration of the larger car is 20.0% greater than the magnitude
of the acceleration of the smaller. Determine the magnitude of acceleration for each car
such that when their front bumpers touch, they have just come to rest.
Step 1. Understanding the Problem
a) Translation
Quantities with subscript s will refer to the smaller car, while quantities with subscript
l refer to the larger car. The vectors ~x, ~v , and ~a, with magnitudes x, v, and a, will
represent displacement, velocity, and acceleration, respectively. I’ll choose coördinates
so that the smaller car is at the origin and the larger car is on the positive x-axis. To save
time, I’ll use d to represent the initial distance between the two cars (i.e., d = 275 m)
and f to represent the ratio of their accelerations (i.e., al = f as ).
We are given as initial data that
~xsi
~xli
~vsi
~vli
~al
= ~0 m
= dx̂
= (25.0 m/s)x̂
= (−30.0 m/s)x̂ = −f~vsi
= −1.2~as = −f~as .
I’ll note the coïncidence that vli = f vsi ; this may make the final algebra simpler.
At the moment they collide, we have xsf = xlf and we are told that vsf = vlf = 0 m/s.
We are asked to find as and al .
b) Laws, Assumptions, Simplifications
This is a one-dimensional problem, so we can simplify the problem by working directly
with the x-components of our vectors, which will be just be either positive or negative
their magnitudes, depending on direction.
We assume that the acceleration of each car is constant. In particular, we can apply
kinematic relationships to the problem.
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PH 221
Homework 2 Solution
Winter 2017
Step 2. Representing the Problem
a) Physical Representation
d = 275 m
as =?
vsi = 25.0 m/s
vli = 30.0 m/s
al = 1.2as
b) Mathematical Representation
vf 2 = vi 2 + 2a∆x
Step 3. Solving the Problem
From the problem statement, we have xsi = 0 m, xli = d, and xsf = xlf , which implies
∆xs = xsf − xsi = xsf
∆xl = xlf − xsi = xsf − d
We have also that vsf = vlf = 0, al = f as , and vli = f vsi . Inserting all of this into the
mathematical representation from Part (2b) for both cars, we get
0 = vsi 2 − 2as xsf
2
0 = vli + 2f as xsf
(1)
− d = f 2 vsi 2 + 2f as xsf − d ,
(2)
where the minus-sign in the Equation (1) is due to the fact that the acceleration of the
small car is in the negative x-direction. We can solve Equation (1) to find xsf = 12 vsi 2 /as .
Inserting this result into Equation (2) yields
2
vsi
2
2
0 = f vsi + 2f as
−d
2as
= f 2 vsi 2 + f vsi 2 − 2f das .
(3)
Finally, we solve Equation (3) for as , finding
as =
(1 + f )vsi 2
(2.2) (25.0 m/s)2
=
= 2.50 m/s2 .
2d
2(275 m)
It follows that
al = 1.2as = 3.00 m/s2 .
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As in your
textbook, these
right-justified
numbers in
parentheses are
“equation
numbers”.
They allow us
to reference
equations “by
name” in the
main text.
PH 221
Homework 2 Solution
Winter 2017
Step 4. Reflecting on the Solution
The solution makes sense.
The units cancel appropriately to give m/s2 , as expected for accelerations.
Both values are positive, as expected for magnitudes.
The values are neither unnaturally large nor unnaturally small for a physically realistic
system.
We can check our answer in a number of ways. One method is to take our acceleration
and known initial velocity for each car and compute how far it travels before coming to
rest. Doing this, we find that the small car travels 125 m while the large car travels 150 m,
which, as expected, corresponds to a total distance of 275 m. We can, alternatively, use
the other kinematic equations. For example, we know vf = vi + a∆t. Plugging in our
solution and initial data for the two cars, we find that the stopping time in both cases
is 10 s. Inserting this into the general equation ∆x = vi ∆t + 12 a(∆t)2 yields the same
displacements of 125 m and 150 m.
Considering our symbolic expression for as , we see that it would get smaller if we were
to increase d, which is expected: if the cars have a longer distance over which to stop,
they need not accelerate so greatly.
Conversely, increasing the initial speed of the car would cause our as to increase, accounting for the necessity of shedding additional speed while covering the same distance.
We can also consider the effect of limiting cases on f . In the special case f = 1, the
solution is precisely what you would find for a car that must stop in a distance d/2.
This is to be expected, because f = 1 means the cars have the same initial speed and
the same acceleration, so, of course, they travel the same distance. Alternatively, we
can consider the case f = 0, which means that the larger car is initialy at rest and has
zero acceleration. In this case we find that our expression is precisely that of a car that
travels a distance d before coming to rest, which of course the second car would need to
do in order to satisfy the problem statement.
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