These formulas are for constant acceleration along one dimension.

(1) x = x_i + v_i * t + (1/2)at²

(2) v = v_i + at,

where * means times,

x_i is the initial position, x is the position at time t,

v_i is the initial velocity, v is the velocity at time t,

and a is the (constant) acceleration.

I type ^2 to mean "squared".

From these it can be shown that:

(3) x - x_i = (1/2) ( v_i + v)t, and

(4) v² = v_i² + 2a(x - x_i).

Comments

-Many textbooks call formulas (1) through (4) called the "kinematic equations".

-These equations can be written in many slightly different ways. For example, many books write v as v_f for "final velocity". Some books write x_o and v_o instead of x_i and v_i. Also, x - x_i; is the displacement, which is often written as d.

-Here is a review of the definitions of acceleration and velocity.

Some solved problems

Problems are roughly in increasing order of difficulty.

Problem 1) An SUV with an initial velocity of 20.0 m/s accelerates at 2.00 m/s^2 for 5.00 seconds. What is its final velocity?

Solution This can be done with equation (2). We have v_i = 20.0 m/s, a = 2.00 m/s^2 and t = 5.00 seconds. Equation (2) is:

v = v_i + at.

Plugging in, v = 20.0 m/s + ( 2.00 m/s^2)( 5.00 s) = 30.0 m/s

Problem 2) A stone is dropped from a bridge. After 1.5 seconds it hits the ground. How far did it fall?.

Solution This can be done using equation (1). We know that t is 1.50 s. If the stone was "dropped", we can say that v_i was about 0.

We need a value for the acceleration of the stone. If we can ignore air friction, we can say that the acceleration of the stone is constant, and equal to g, the acceleration due to gravity. g varies slightly from place to place on the surface of the earth, but a typical value of g would be about 9.80 m/s^2. If we take the positive direction to be downwards, then we can say that a is +9.80 m/s^2.

So x - x_i = v_i * t + (1/2)at² = 0 * (1.50 s) + ( 1/2)(9.80 m/s^2)(1.50 s)^2 = 11.025 m or about 11.0 m .

Problem 3) A car, starting at rest, accelerates at a constant acceleration, and is moving at 35 m/s after 10 seconds. How far did it go in those 10 seconds?

Since the car "started at rest" we can say v_i = 0. The final velocity v = 35 m/s. The time t = 10 s.

First solution Probably the fastest way to solve this is using equation (3). The displacement d is:

d = x - x_i = (1/2) ( v_i + v)t = (1/2)(0 + 35 m/s)(10 s) = 175 m

Second solution We can use equation (2) to find the acceleration a and then use equation (1) to find the displacement.

Equation (2) is: v = v_i + at

So 35 m/s = 0 + a(10 s)

Solving this gives a = ( 35 m/s) / ( 10 s) = 3.5 m/s.

Then equation (1) is: x = x_i + v_i * t + (1/2)at²

From equation (1), x = x_i + v_i * t + (1/2)at² = x_i + 0( 10 s) + (1/2)(3.5 m/s^2)(10 s)^2

or x = +175 m.

So d = x - x_i = 175 m, which is of course the same answer we got before. This is an example of how there is often more than one way to do a problem involving constant acceleration.

Problem 4) "A sports car moving at 30 m/s needs to stop in a distance of 100 m. If its accleration while braking is constant, what will the magnitude of that acceleration be?"

Solution The initial velocity v_i is 30 m/s. The velocity when the car has stopped will be 0, so v = 0. The displacement d = x - x_i is 100 m.

The shortest way to solve this is using (4).

v² = v_i² + 2a(x - x_i).

Solving this for a gives:

a = [v² - v_i²] / (x - x_i). = [(0 m/s)^2 - (30 m/s)^2] / (100 m) = -4.5 (m^2/ s^2) / m = -4.5 m/s^2

Where did the minus sign come from? At the start, when we said v_i = + 30 m/s, we took the positive direction to be the direction in which the sports car was moving. Since the car is braking (slowing down), its acceleration is opposite to the direction in which it is moving, and thus is negative.

Now the magnitude of the acceleration a is the magnitude of -4.5 m/s^2, which is just 4.5 m/s^2.

Problem 5) A basketball player throws a ball straight upwards at a velocity of 6.67 m/s from a height of 2.50 m above the floor. How many seconds will it be until the ball hits the floor?

Solution If we take the positive direction to be up, we can say v_i is +6.67 m/s. Then if we take the zero of height to be the floor we can say x_i = 2.50 m. When the ball hits the floor, its height x will be zero.

We can solve this using equation (1) again, but this time we will get a quadratic equation.

x = x_i + v_i * t + (1/2)at²

Subtracting x from both sides, and rearranging a little bit, /DIV>

(1/2)at² + v_i * t + x_i - x = 0

This is a quadratic equation in time t.

Putting the numbers in,

(1/2)(-9.80 m/s^2)t² + (6.67 m/s)t + 2.5 m = 0

The accleration is negative, because things fall downwards, and we chose the positive direction to be up. We can multiply both sides of this equation by -1 to get

(1/2)(9.80 m/s^2)t² - (6.67 m/s)t - 2.5 m = 0

This is at^2 + bt + c = 0 with a = 4.90 m.s^2, b = -6.67 m/s and c = -2.5 m.

Using the quadratic formula we get that

with Sqrt meaning "the square root of".

This simplifies to:

t = [ 6.67 m/s + /- 9.67 m/s] / (9.80 m/s) .

So t = either 1.67 s or -0.31 s.

The negative solution can't be right, so we choose the t = 1.67 s solution.

Another way to do this is to solve the quadratic equation in letters, and then put the numbers in at the end.

The solution in letters is:

with h = 2.50 m being the height the ball was thrown from.

If we put the numbers in now, we get the same answer as before.

Applications

How are these equations used?

-Constant velocity Motion at constant velocity is the same thing as acceleration of zero, so it is an example of constant acceleration. So, motion with constant velocity can be treated by just putting in 0 for the acceleration a.

-Vehicles. As you will have noticed, physics problems often talk about vehicles moving with constant acceleration. One reason for this is that vehicles do sometimes move at nearly constant acceleration for a limited time.

-Projectiles If you can ignore air friction and pretend that the earth is flat, the vertical motion of a projectile near the earth has a constant downward acceleration. For the motion of projectiles in two dimensions, if you can pretend the earth is flat, and ignore air friction, you can treat the horizontal motion as one dimensional motion with constant velocity, and the vertical motion as one dimensional motion with constant downward acceleration. At times, this can be a good approximation.

-Constant angular acceleration around one axis The equations of motions for constant angular acceleration around one axis are completely analogous to the equations of motion for constant acceleration along one dimension.

-Anything, for a short time. Nothing moves at non-zero acceleration forever. But objects tend to move at approximately constant acceleration for a limited time. If the acceleration of an object changes with time, the time interval can often be divided up into smaller time intervals, and the acceleration treated as being constant over each of the smaller time intervals.