Associated Topics || Dr. Math Home || Search Dr. Math

### Newton's Laws Applied to an Accelerating Car

```Date: 05/26/2005 at 01:09:58
From: Pun
Subject: Car acceleration Newton laws

When a car accelerates, what external force is acting on it?  When a
person sits in the car, why does he feel "backwards movements" on him?
My teacher said it's something to do with inertia of the body and
Newton's Law 1 and Newton's Law 3.  Do you feel a forward or backwards
movement first?

Can you explain every single aspect when the car accelerates until
the person is moving in the same state of motion as the car?

```

```
Date: 05/26/2005 at 08:30:30
From: Doctor Rick
Subject: Re: Car acceleration Newton laws

Hi, Pun.

When a car accelerates, it is because the road is pushing it forward!

Here's a little more detail.  The car's engine turns the wheels.  When
the tire turns at a constant rate, the point at the bottom of the tire
is stationary for a moment; the wheel acts like a lever with the
fulcrum at the bottom, because friction of the tire against the road
keeps it from slipping.  The axle, at the center of the wheel, moves
at the speed of the car; the point at the top of the wheel moves twice
as fast.

When the car is accelerating, the wheels undergo angular or rotational
acceleration: they turn faster and faster.  This is accomplished by
angular force (called torque) exerted by the engine on the axle.  At
the bottom of the tire, this torque takes the form of a force acting
backward, pushing against the road.  But as long as the tire is not
slipping (skidding), the point at the bottom of the tire continues to
be stationary: it is not accelerating.  By Newton's first law, it is
"an object at rest and staying at rest" and this is only possible if
the net force on it is zero.

The axle is exerting a force on that point, pushing backward against
the road.  In order for the net force to be zero, there must be
another force of equal magnitude, pushing forward on the point at the
bottom of the tire.  This force is exerted by the road; it is Newton's
third law "equal and opposite" reaction force.

When we turn our attention from the point at the bottom of the tire to
the entire car, we find that this reaction force of the road against
the tire, pushing forward, is the only unbalanced force acting on the
car.  (I'm assuming a flat road, so that the force of gravity pulling
down on the car is balanced by the "normal force" of the road pushing
up on the car, keeping the car from sinking into the road.)  When an
unbalanced force acts on an object (the car), the object accelerates
in the direction of the force (Newton's second law).  So it's the road
pushing on the car that makes it accelerate forward.  Of course the
power comes from the car's engine, but in terms of external forces,
that's the way it is!

Now we can turn our attention to you, sitting there with your seat
belt on.  As the car moves, your body is subject to Newton's first
law: it is minding its own business, moving along at constant speed
(let's say 60 miles per hour) along with the car.  But as the car
accelerates, the next moment the seat back is moving slightly faster
than you are, so it catches up with you.  You and the seat back can't
be in the same place at the same time, so the seatback pushes forward
against you: it exerts a force directed forward.  By Newton's third
law, your body exerts a reaction force against the seat back,
directed backward.

Let's get some math into the picture.  Let's say the road is exerting
a forward-pointing force F against the car, the car (without you) has
mass M and its acceleration is A.  You have mass m, and the car is
pushing against you with force f.  Your body pushes back against the
seat back with equal force f, and it accelerates with acceleration a.
I can set up two free-body diagrams, one for the car and one for you:

f<---- CAR ---->F          YOU ---->f
MA                  ma

By Newton's second law, I derive the following equations from the
free-body diagrams:

F-f = MA
f = ma

I use the second equation to substitute into the first equation:

F - ma = MA

I'd need another equation in order to solve for both accelerations, a
and A.  This equation would have to do with how hard your body and the
seat back push back when something gets too close.  It won't be a
simple equation, but I can assure you that eventually the seat back
must stop getting closer to you: its acceleration and yours become
equal.  Once that has happened, we have

F - ma = Ma
F = Ma + ma
F = (M+m)a
a = F/(M+m)

This is no surprise, because when we consider the car plus you to be
a single free body, we get the diagram

CAR+YOU ----> F
(M+m)a

All the time the car continues accelerating with constant acceleration
a, you will feel that force of the seat pushing forward against you,
and you will be pushing backward against the seat.  If you put a
bathroom scale between you and the seat back, it would register a
weight.  The scale normally measures the force with which the earth
pushes up on you to keep you from falling through; in this case it is
measuring the force with which the seat back pushes forward on you to
get you going faster.  That's why when the car accelerates, you feel
as if you are pushed back into the seat.

One more step: Let's say the car has reached 65 miles per hour, which
is the speed limit, so it stops accelerating and goes at a steady 65
miles per hour.  Now there is no more force between you and the seat.
But when the car turns off the highway and decelerates to 45 miles per
hour, the forces reverse direction.  Now your body is moving faster
than the car, so it moves forward until the seat belt stops it.  In
order to stop you from moving forward, the seat belt exerts a force on
you, and you exert an equal and opposite forward force against the
seat belt.  When the car gets down to 45 miles per hour and stops
decelerating, the forces go away again.

If you'd like to discuss a point further, I'll be happy to do so.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search