Constant Acceleration Motion

Constant acceleration motion is motion for which the acceleration remains unchanged during the time interval being studied. In one dimension,

(1)

This simple condition produces one of the most important families of equations in introductory mechanics. Because acceleration is the time derivative of velocity, and velocity is the time derivative of position, the constant acceleration equations follow directly by integration.

The results are not independent formulas to memorize. They are different algebraic forms of the same underlying kinematics.

1 Assumptions and scope

For one-dimensional motion, the constant acceleration model assumes

(2)

where a does not vary with time over the interval of interest.

The model applies exactly to idealized cases such as a particle moving under a constant net force with constant mass. It also provides an excellent approximation for many short-duration problems in which the acceleration changes negligibly.

It does not apply without modification when acceleration depends appreciably on time, position, or velocity. Examples include strong aerodynamic drag, a spring force over a large interval, or gravity over distances large enough that g changes significantly.

2 Derivation of the velocity equation

Acceleration is defined by

(3)

For constant a,

(4)

Integrate from the initial state at t = 0, where the velocity is v_0x, to a later time t:

(5)

Therefore

(6)

or

(7)

The velocity therefore changes linearly with time. On a velocity-time graph, the slope is the constant acceleration.

3 Derivation of the position equation

Velocity is

(8)

Insert the constant acceleration velocity law:

(9)

Thus

(10)

Integrating from x₀ at t = 0 to x at time t gives

(11)

Hence

(12)

Using the displacement Δx = x − x₀,

(13)

Because the position contains a term proportional to t², the position-time graph is parabolic when acceleration is nonzero.

4 Average velocity for constant acceleration

When acceleration is constant, the velocity changes linearly with time. The average value of a linear function over an interval is the arithmetic mean of its endpoint values. Therefore

(14)

Since displacement equals average velocity times elapsed time,

(15)

This result is specific to constant acceleration. It is not generally true when acceleration varies with time.

5 Eliminating time

Some problems give positions and velocities but not the elapsed time. Time can be eliminated algebraically.

From

(16)

we have

(17)

provided a≠0.

Insert this into

(18)

Then

(19)

Using

(20)

we obtain

(21)

This form is especially useful when time does not appear among the known or requested quantities.

6 The four standard one-dimensional relations

For constant acceleration in one dimension, the most useful equations are

(22)

(23)

(24)

and

(25)

These equations contain the same physical information. Which form is most convenient depends on which variables are known.

7 Sign conventions

The equations are vector-consistent only when signs are assigned according to a chosen positive direction.

For one-dimensional motion, choose the positive axis first. Then:

• velocity is positive when motion is in the positive direction;
• velocity is negative when motion is in the negative direction;
• acceleration is positive when the acceleration vector points in the positive direction;
• acceleration is negative when it points in the negative direction;
• displacement is positive or negative according to the final position relative to the initial position.

A negative acceleration does not automatically mean that an object is slowing down. An object speeds up when velocity and acceleration have the same sign and slows down when they have opposite signs.

8 Vertical motion near Earth’s surface

When aerodynamic drag is neglected and the vertical range is small compared with Earth’s radius, gravitational acceleration near the surface is approximately constant.

Choose upward as positive. Then

(26)

where

(27)

The constant acceleration equations become

(28)

(29)

and

(30)

At the highest point of an upward throw,

(31)

but the acceleration is still

(32)

A downward-positive coordinate system is equally valid. In that convention a_y = +g. The physics is unchanged as long as one convention is used consistently.

9 Vector form

For motion in several dimensions with a constant acceleration vector a, the one-dimensional relations generalize componentwise:

(33)

and

(34)

Thus each Cartesian component obeys its own constant acceleration equation. Projectile motion is an important application: with negligible air resistance, the horizontal acceleration is approximately zero and the vertical acceleration is approximately −g.

10 Dimensional checks

Every term in a kinematic equation must have the same dimensions.

For

(35)

we have

(36)

and

(37)

Similarly, in

(38)

both v² and aΔx have dimension L²∕T².

Dimensional consistency cannot prove that an equation is correct, but it can quickly expose many algebraic errors.

11 Worked example 1: braking to rest

A CAR travels at

(39)

and brakes with constant acceleration

(40)

Find the stopping time and stopping distance.

Choose the initial direction of motion as positive. At rest,

(41)

Using

(42)

we get

(43)

so

(44)

The displacement is

(45)

which gives

(46)

The negative acceleration describes braking because the velocity is initially positive.

12 Worked example 2: ball thrown vertically upward

A ball is launched upward from y₀ = 0 with

(47)

Neglect air resistance. Find the time to maximum height, the maximum rise, and the time to return to the launch height.

Choose upward as positive, so

(48)

At the top,

(49)

Therefore

(50)

so

(51)

Use the no-time equation for the rise:

(52)

Hence

(53)

Because the ball returns to its launch height under constant downward acceleration with no drag, the ascent and descent times are equal. Thus

(54)

At that instant the velocity is approximately −18.0 m/s. The speed magnitude equals the launch speed, but the direction is downward.

13 Worked example 3: two-dimensional constant acceleration

A particle has

(55)

(56)

and constant acceleration

(57)

Find the velocity and position at t = 3.0 s.

The velocity is

(58)

so

(59)

or

(60)

The position is

(61)

Thus

(62)

The displacement is therefore

(63)

with magnitude

(64)

14 Worked example 4: find acceleration without finding time first

A vehicle speeds up uniformly from

(65)

to

(66)

over a displacement of

(67)

Find the acceleration and elapsed time.

Because time is not initially known, use

(68)

Then

(69)

so

(70)

Now use

(71)

to obtain

(72)

This example illustrates why equation selection should follow the available variables rather than a fixed memorized order.

15 Common mistakes

• Using the constant acceleration equations when acceleration actually varies significantly.
• Substituting the magnitude g without first deciding whether it is positive or negative in the chosen coordinate system.
• Assuming negative acceleration always means slowing down.
• Assuming that v = 0 implies a = 0 at the top of vertical motion.
• Mixing position x with displacement Δx = x − x₀.
• Using Δx = (v₀ + v)t∕2 when the acceleration is not constant.
• Forgetting that the equation v² = v ₀² + 2aΔx loses the sign of v when one later takes a square root.

16 Practice problems

M01-05-P01

A particle begins with v₀ = 3.0 m∕s and has constant acceleration a = 2.0 m∕s². Find its velocity after 5.0 s.

M01-05-P02

A car starts from rest and accelerates at 3.0 m∕s² for 6.0 s. Find its displacement.

M01-05-P03

A cyclist moving at 12 m∕s slows uniformly at −2.0 m∕s². Find the time required to reach 4.0 m∕s.

M01-05-P04

A train speeds up uniformly from 8 m∕s to 20 m∕s in 6.0 s. Find its acceleration and displacement.

M01-05-P05

A car traveling at 30 m∕s brakes uniformly at −5.0 m∕s². Find the stopping distance.

M01-05-P06

A stone is dropped from rest. Neglect air resistance and take downward as positive. Find its speed and downward displacement after 2.0 s using g = 9.81 m∕s².

M01-05-P07

A ball is thrown straight upward at 14 m∕s. Take upward as positive. Find the time to the highest point and the maximum rise.

M01-05-P08

A particle has x₀ = −4 m, v₀ = 5 m∕s, and a = −1.5 m∕s². Find x and v at t = 4.0 s.

M01-05-P09

A vehicle covers 72 m while accelerating uniformly from 6 m∕s to 18 m∕s. Find the acceleration.

M01-05-P10

A particle moving at −10 m∕s has acceleration −2.0 m∕s² for 3.0 s. Is it speeding up or slowing down? Find its final velocity.

M01-05-P11

A particle has v₀ = 2e_x + 3e_y m∕s and a = 4e_x − 2e_y m∕s². Find v after 2.0 s.

M01-05-P12

A ball is released from rest from a height of 19.6 m above the ground. Neglect air resistance and use g = 9.8 m∕s². Find the time to reach the ground and the speed immediately before impact.

17 Compact answer check

1. 13 m∕s.
2. 54 m.
3. 4.0 s.
4. 2.0 m∕s² and 84 m.
5. 90 m.
6. 19.6 m∕s and 19.6 m downward.
7. 1.43 s and approximately 10.0 m.
8. x = 4 m and v = −1 m∕s.
9. 2.0 m∕s².
10. Speeding up; v = −16 m∕s.
11. v = 10e_x − 1e_y m∕s.
12. 2.0 s and 19.6 m∕s.

18 Connection to the next article

Constant acceleration is special because integration gives simple polynomial functions of time. M01-06 removes that restriction and studies variable acceleration, for which one must work directly with

(73)

and

(74)

or with position-dependent and velocity-dependent forms of acceleration.

References