 This article is the second chapter in a series on how to understand and approach kinematics problems. The first chapter covered position, velocity, and acceleration. Now that we understand these quantities, we are going to use them to solve problems in one dimension.

# Kinematics Equations for Constant Acceleration

The four horsemen of the kinematics apocalypse are:

xf – xi = (vf – vi)*t/2

vf – vi = a*t

vf2 = vi2 + 2*a*(xf – xi)

xf = xi + vi*t + ½a*t2

Note: the little f stands for final (as in the final velocity or position) while the little i stands for initial.

Note: These equations only work for constant acceleration, but nearly all problems have constant acceleration.

# 1-Dimensional Problem Solving Steps

For every one dimensional kinematics problem, the steps are pretty much the same.

• Write down every quantity the problem gives you (initial and final position, initial and final velocity, acceleration, time, etc)
• Write down which quantity you are trying to find
• Find the kinematic equation (or sometimes two equations) to relate these quantities.
• Solve the algebra.

Yes, it really is that simple. (In fact, most physics problems work the same way. For more details on the physics problem solving algorithm, check out this article.)

## Avoiding Common Mistakes: Hidden Quantities

Sometimes the problem may tell you a quantity secretly; you may not even realize you got it. For example, if they tell you the displacement (how far something travelled) but not positions, you can treat the displacement as xf and set xi to 0. Likewise, if the problem doesn’t say anything special about acceleration, then the acceleration is probably just gravity, a = g = 9.8m/s2. These hidden quantities are as valid as regular quantities, they are just a little harder to spot.

## Avoiding Common Mistakes: Top of Flight

A special example of the hidden quantity is when they tell you an object is at “the top of its flight/motion/path/etc.” This means they are secretly telling you that vf is 0 because an object traveling in one dimension always has a velocity of 0 at the top of its path. Wondering why? Well, if the velocity was going up, then a millisecond later, the object would be higher (and thus it can’t be at the top of its path). Likewise, if the object had a velocity downward, then it would have been higher a millisecond before (so it can’t be at the top either).

## Avoiding Common Mistakes: Positive and Negative Numbers

It can be tricky to keep track of your negatives. The key is direction; down is always negative. So if an object is going down, it will have a negative velocity. If the acceleration is going down (which it almost always is) then the acceleration is negative. And don’t forget from our first chapter, it is possible to have positive velocity and negative acceleration at the same time!

## Example: A Woman and Her Ball

A woman is holding a ball at 1 meter, and throws it upward at 5m/s. a) How high does the ball reach? b) How long does the ball take to hit the ground? c) How fast is the ball going when it hits the ground?

Let’s find out!

Part A: How high does the ball reach?

What do we know?

The initial position xi = 1 m
The initial velocity vi = 5 m/s
Secret Quantity: a = -9.8 m/s2 (gravity)
Secret Quantity: At the top of the ball’s arc (i.e. when its at its highest) vf = 0 m/s

What are we trying to find?

The position at the top of the throw, xf

What equation relates these quantities?

We’re looking for an equation that includes xf, xi, vf, vi, and a
vf2 = vi2 + 2*a*(xf – xi) seems to fit nicely!

Plug in and solve

vf2 = vi2 + 2*a*(xf – xi)
(0 m/s)2 = (5 m/s)2 + 2*(-9.8 m/s2)*(xf – 1m)
0 = 25 (m2/s2) – (19.6 m/s2)*(xf – 1m)
(19.6 m/s2)*(xf – 1m) = 25 m2/s2
xf –1m = (25 m2/s2)/(19.6 m/s2)
xf –1m = 1.28 m
xf = 2.28 m

Tada! The final height = 2.28 meters

Part B: How long does the ball take to hit the ground?

What do we know?

We still know xi = 1 m, vi = 5 m/s and a = -9.8 m/s2 but now we also know that xf = 0 m (because the height of the ground is 0m)

Note: Since the ball is now on the ground instead of at the top of it’s flight, vf ≠ 0 so that is off the table.

What are we trying to find?

The time, t

What equation relates these quantities?

We’re looking for an equation that includes xf, xi, vi, a and t
Looks like we’re going to need xf = xi + vi*t + ½a*t2

Plug in and solve

xf = xi + vi*t + ½a*t2
0 m = 1 m + (5 m/s)*t + ½ (-9.8 m/s2)* t2
-(4.9 m/s2)* t2 + (5 m/s)*t + 1 m = 0

This is a quadratic equation (ax2 + bx + c = 0) and can be solved using the quadratic formula.

t =
t =
t =  or t =
t =  or t =
t = -.17 s or 1.2 s

We can ignore t=-.17 because we are not allowed to have a negative time (we call this a non-physical answer), which leaves us with time = 1.2 seconds!

Part C)

What do we know?

We still know xi = 1 m, vi = 5 m/s, a = -9.8 m/s2 and xf = 0 m but now we also know t = 1.2 seconds because we just solved it.

What are we trying to find?

The velocity, vf

What equation relates these quantities?

We have so many quantities we could use any of the equations, but let’s go with vf - vi = a*t because it’s simple and we haven’t used it yet.

Plug in and solve

vf – vi = a*t
vf – 5 m/s = (-9.8 m/s2) * 1.2 s
vf = 5 m/s – 11.8 m/s
vf = -6.8 m/s

Boom, the velocity when the ball hits the ground is 6.8 m/s downward (thus the negative sign)

# Conclusion

We now know how to use our kinematics equations to solve problems in one dimension. In our next blog, we’ll discuss how to “butcher” vectors in preparation for Kinematics in 2-Dimensions.

Stayed Tuned! Same bat time, same bat channel.

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