Equations of Motion

These equations give us the average velocity and acceleration. The do not tell us the exact value of velocity or acceleration, just an average.

$$v_{\mathrm{avg}} = \frac{\Delta x}{\Delta t} \ \ \ \ \ a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t} $$

The Equations of Motion

If an object's acceleration is constant the equations of motion are also true. They give exact values of initial and final velocity, not averages. Here is the derivation of these equations.

$$v = u+a \Delta t$$ $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = \tfrac{1}{2}(v+u)\Delta t$$ $$v^{2} = u^{2}+2a \Delta x$$ $$\Delta x = v\Delta t - \tfrac{1}{2}a \Delta t^{2}$$

\(\Delta x\) = distance [m] vector

\(\Delta t\) = time period [s] vector

\(v\) = final velocity [m/s] vector

\(u\) = initial velocity [m/s] vector

\(a\) = acceleration [m/s²] (constant) vector

Example: A bullet aimed straight up leaves the barrel of a gun at 400 m/s. It accelerates down at 9.8 m/s². If the bullet travels for 40.8 s before stopping how far up did it go?
(In the real world a bullet would stop quicker because of air friction.)
solution $$\text{list known and unknown variables}$$ $$u = 400 \, \mathrm{\tfrac{m}{s}}$$ $$a = -9.8 \, \mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t = 40.8 \, \mathrm{s}$$ $$\Delta x =\, ?$$
$$\text{identify matching equation}$$ $$\Delta x = u\Delta t + \tfrac{1}{2} a \Delta t^{2}$$
$$\text{plug in values}$$ $$\Delta x = (400) (40.8) + \tfrac{1}{2} (-9.8)(40.8^{2})$$ $$\Delta x = 16320 - 8157$$ $$\Delta x = 8163 \, \mathrm{m}$$

Example: A car moving at 30 m/s puts on its brakes and comes to a stop in 10 meters. What is the acceleration of the car?
solution $$\text{list known and unknown variables}$$ $$u = 30 \, \mathrm{\tfrac{m}{s} }$$ $$v = 0 \, \mathrm{\tfrac{m}{s}}$$ $$\Delta x = 10 \, \mathrm{m}$$ $$a =\, ?$$
$$\text{identify matching equation}$$ $$v^{2} = u^{2}+2a \Delta x$$
$$\text{solve for unknown}$$ $$v^{2} - u^{2}=2a \Delta x$$ $$\frac{v^{2} - u^{2}}{2 \Delta x}=a$$
$$\text{plug in values}$$ $$\frac{0^{2} - (30 \, \tfrac{m}{s})^{2}}{2 (10 \, \mathrm{m})}=a$$ $$\frac{-900 \, \mathrm{\tfrac{m^2}{s^2}} }{20 \, \mathrm{m}}=a$$ $$-45 \mathrm{\tfrac{m}{s^{2}}}=a$$
Example: A plane takes off at a speed of 170 miles/hour while accelerating from rest on a runway that is 6,000 ft long. Find the acceleration of the plane in m/s²?
unit conversion: 1m = 3.3 ft , 1 mile = 1609 m
solution $$\text{convert units to metric}$$ $$\begin{aligned} v &= \mathrm{170\left(\frac{\color{red}{mile}}{\color{blue}{hour}}\right)\left(\frac{1609 \, m}{1 \, \color{red}{mile}}\right)\left(\frac{1 \, \color{blue}{hour}}{3600\,s}\right) = 76\,\mathrm{\tfrac{m}{s}} } \\ u &= \mathrm{rest} = 0 \\ \Delta &x = 6000\,\mathrm{ft}\scriptsize \left(\frac{1 \, \mathrm{m}}{3.3 \, \mathrm{ft} }\right)\normalsize = 1818 \, \mathrm{m} \\ a &= \,? \end{aligned}$$
$$\text{plug variables into matching equation}$$ $$v^{2} = u^{2}+2a \Delta x$$ $$v^{2} - u^{2}=2a \Delta x$$ $$\frac{v^{2} - u^{2}}{2 \Delta x}=a$$ $$\frac{76^{2} - (0^{2})}{2 (1818)}=a$$ $$\frac{5776}{3636}=a$$ $$1.589 \, \mathrm{\tfrac{m}{s^{2}} }=a$$

Acceleration of Gravity

On earth everything is pulled down at 9.8 m/s². Generally the letter g is used for this number. The rate is the same for cars, birds, puppies, apples, balloons, and everything. Other effects like air friction, thrust, and buoyancy can change the precived acceleration. Acceleration from gravity also changes based on the mass of the planet and your distance to the center of the planet.

g = acceleration from gravity on earth = 9.8 m/s²

What is the acceleration in m/s² on the surface of Mars?
(check Wolfram alpha)


What is the acceleration in m/s² on the surface of the Moon?
(check Wolfram alpha)

Example: A sleeping cat falls from rest off of a ledge. If the cat hits the ground moving at 6.0 m/s how long was the cat in free fall?
solution
  • $$u = \mathrm{rest} = 0$$ $$v = -6.0 \, \mathrm{\tfrac{m}{s}} $$ $$a = -9.8 \,\mathrm{ \tfrac{m}{s^{2}} }$$ $$\Delta t = \,? $$
  • $$v = u+a \Delta t$$ $$\frac{v - u}{a} = \Delta t $$ $$\frac{(-6\, \mathrm{\tfrac{m}{s} }) -(0)}{-9.8 \, \mathrm{\tfrac{m}{s^{2}}}}= \Delta t $$ $$0.61\,\mathrm{s} = \Delta t$$
  • Example: A meteor 4,000 m above Europa, a moon of Jupiter, falls from rest. What is the impact velocity of the meteor? How long does it take for the meteor to hit the surface? (look up the gravity on Europa)
    solution
  • $$\Delta x = -4000\,\mathrm{m}$$ $$u = \mathrm{rest} = 0$$ $$a = -1.315\, \mathrm{\tfrac{m}{s^{2}}}$$ $$v = \,?$$ $$\Delta t = \,?$$
  • $$v^{2} = u^{2}+2a \Delta x$$ $$v^{2} = 0^{2}+2(-1.315)(-4000)$$ $$v^{2} = 10520 $$ $$v = \pm102.6 \, \mathrm{\tfrac{m}{s}} $$

  • $$v = u+a \Delta t$$ $$\frac{v - u}{a} = \Delta t$$ $$ \frac{-102.6 - 0}{-1.315} = \Delta t $$ $$78.0\, \mathrm{s} = \Delta t $$

    2-D Motion

    Solving for 2-Dimensional motion can reuse the methods from 1-Dimension. We just have to divide the problem up into x and y information. The x information is typically horizontal and the y information is typically vertical. As long as x and y are perpendicular to each other the events in each direction don't affect each other. The x and y axises are considered independent from each other.

    In this simulation you can see what a graph of position, velocity, and acceleration looks like for 2-D motion.

    It helps to organize our information in an x column and y column with time separate. The x and y columns are completely independent of each other, but time is shared.

    Δt
    Δx Δy
    u u
    v v
    a a

    If an object is in free fall we know the gravitational acceleration is purely in the vertical and there is zero horizontal acceleration.

    Play around with this 2-D motion solver to get a feel for what paths an object can take. You can also plug in vlaues to see if you did a 2-D motion problem correctly.

    See the Pen ballistics calculator (vue) by Landgreen ( @lilgreenland) on CodePen.

    Example: A ball moving horizontally at 2 m/s rolls off a table that is 1.5 m high. Find how far the ball travels horizontally before it hits the ground. Click to rewatch the simulation above.
    solution

    The information given is organized in the chart below. Vectors directed down or left are negative and vectors directed up or right are positive.

    Δt =
    Δx = ? Δy = -1.5 m
    u = 2 m/s u = 0
    v = 2 m/s v =
    a = 0 a = -9.8 m/s²

    There isn't enough information to solve the horizontal side. Let start on the vertical.

    $$v^{2} = u^{2}+2a \Delta y$$ $$v^{2} = 0^{2}+2(-9.8)(-1.5)$$ $$v^{2} = 24.9$$ $$v = \sqrt{24.9}$$ $$v = \pm4.98 $$ $$v = -4.98 \, \mathrm{\tfrac{m}{s}}$$

    We know the final velocity is down so after the square root we choose the negative value. Lets plug final velocity back in and solve for time.


    $$v = u + a \Delta t$$ $$-4.98 = 0+(-9.8)\Delta t$$ $$0.51 \, \mathrm{s} = \Delta t$$

    With this new information we can use time to solve on the horizontal side.

    Δt = 0.51 s
    Δx = ? Δy = 1.5 m
    u = 2 m/s u = 0
    v = 2 m/s v = -4.98 m/s
    a = 0 a = 9.8 m/s²
    $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = 2(0.51) + \tfrac{1}{2}(0)(0.51)^{2}$$ $$\Delta x = 1.02 \, \mathrm{m}$$
    Example: A pokéball is thrown at an angle of 60° above the horizon and a speed of 10 m/s. If the ball is thrown from 2.0 meters above the ground how far does the ball go before it misses the charmander and hits the ground.
    vector review
    solution
    Δt = ?
    Δx = ? Δy = -2.0 m
    u = 10 cos(60) = 5.0 m/s u = 10 sin(60) = 8.66 m/s
    v = ? v = ?
    a = 0 a = -9.8 m/s²
    $$v^{2} = u^{2}+2a \Delta y$$ $$v^{2} = 8.66^{2}+2(-9.8)(-2.0)$$ $$v^{2} = 114$$ $$v = \pm 10.7 \, \mathrm{\tfrac{m}{s}}$$ $$v = -10.7 \, \mathrm{\tfrac{m}{s}}$$
    $$v = u+a \Delta t$$ $$-10.7 = 8.66+(-9.8)\Delta t$$ $$1.97 \, \mathrm{s} = \Delta t$$
    $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = 5(1.97) + \tfrac{1}{2}(0)(1.97)^{2}$$ $$\Delta x = 9.85 \, \mathrm{m}$$
    Δt = 1.97 s
    Δx = 9.85 Δy = -2.0 m
    u = 10 cos(60) = 5.0 m/s u = 10 sin(60) = 8.66 m/s
    v = 5.0 m/s v = -10.7 m/s
    a = 0 a = -9.8 m/s²
    Δx v θ Example: An arrow is fired from ground level 60° above the horizon at 10 m/s. Find the horizontal distance the arrow traveled.
    solution

    We know that after the arrow goes up it will land at the same height it started at. This means Δy is 0.

    We can also use trig identities to change the initial velocity into x and y parts.

    $$u_x = 10 \mathrm{cos}(60)$$ $$u_x = (10) (0.5)$$ $$u_x = 5 \, \mathrm{\tfrac{m}{s} }$$
    $$u_y = 10 \mathrm{sin}(60)$$ $$u_y = (10) (0.87)$$ $$u_y = 8.7 \, \mathrm{\tfrac{m}{s} }$$
    Δt
    Δx Δy = 0
    u = 5 m/s u = 8.7 m/s
    v v
    a = 0 a = -9.8 m/s²

    We can now solve for the time period with the vertical information.

    $$\Delta y = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$0 = 8.7 \Delta t + \tfrac{1}{2}(-9.8) \Delta t^{2}$$

    We can separate equations equal to zero into two parts.

    $$0 = \Delta t (8.7 + \tfrac{1}{2}(-9.8) \Delta t)$$
    $$0 = \Delta t $$
    $$0 = 8.7 + \tfrac{1}{2}(-9.8) \Delta t$$ $$-8.7 = -4.9 \Delta t$$ $$1.93 \, \mathrm{s} = \Delta t$$

    We can throw out the time being zero since that doesn't make sense for the problem.

    Δt = 1.93 s
    Δx Δy = 0
    u = 5 m/s u = 8.7 m/s
    v v
    a = 0 a = -9.8 m/s²

    With the time we can now use horizontal information to solve for the horizontal distance.

    $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = (5)(1.93) + \tfrac{1}{2}(0)(1.93)^{2}$$ $$\Delta x = (5)(1.93)$$ $$\Delta x = 9.65 \, \mathrm{m}$$
    Δt = 1.93 s
    Δx = 9.65 m Δy = 0
    u = 5 m/s u = 8.7 m/s
    v v
    a = 0 a = -9.8 m/s²
    V = 150 m/s Vx Vy θ = 35° Example:
    A ball is thrown at 35° from the horizon
    from ground level
    with a speed of 150 m/s.
    How long will the ball be in the air?
    solution
    Δt = ?
    Δx = Δy = 0 m
    u = 150 cos(35) = 123.0 m/s u = 150 sin(35) = 86.0 m/s
    v = v =
    a = 0 a = -9.8 m/s²
    $$\Delta y = u\Delta t + \small\frac{1}{2}a \Delta t^{2}$$ $$0 = (86.0) \Delta t + \small\frac{1}{2}(-9.8)\Delta t^{2}$$

    We don't have to use the quadratic equation if we factor out time and solve two separate equations.

    $$0 = \Delta t(86.0 + \small\frac{1}{2}(-9.8)\Delta t)$$ $$0=\Delta t \quad \quad 0 = 86.0 + \small\frac{1}{2}(-9.8)\Delta t$$

    The solution of zero time isn't helpful.

    $$0 = 86.0 + \small\frac{1}{2}(-9.8)\Delta t$$ $$-86.0 = -4.9\Delta t$$ $$\Delta t = 17.6 \, \mathrm{s}$$
    Challenge: Test your 2-D motion skills with these randomly generated problems.

    2-D motion challenge #1
    2-D motion challenge #2

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