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### Course: High school physics>Unit 1

Lesson 6: Motion with constant acceleration

# Choosing kinematic equations

Kinematic equations help solve for an unknown in a problem when an object has either a constant velocity or constant acceleration. This video will help you choose which kinematic equations you should use, given the type of problem you're working through.

## Want to join the conversation?

• For the first question, don't we have to convert the 80km/h to m/s before calculating?
• Yes. At , there is a note that says that the conversion must be done first, and the instructor messed up. If you didn't convert, at best you'd have a convoluted answer containing a ridiculous fraction with units everywhere.
• So, what are the actual answers to these questions? I solved them myself as well, but I'm new to this and may have done something wrong along the way.

Thanks!
• you will learn it soon enough!

for the first problem x= 182.87m and t=16.46s
for the second problem x=172.8m and final v is 28.8m/s
• How does one derive the 4th formula (v^2 = (v0)^2 +2aΔx)?
• Looking at the equation you notice there is no t. So, we need to use the equations we know and replace t. The easiest one to do that for is the first equation. Then use that information in another equation to get our 4th,

Vf = Vo + at to solve for t, subtract Vo from both sides Vf - Vo = at
now divide both sides by a and we get Vf -Vo/a = t

Now just use equation 3 replacing t with Vf - Vo/a

I don't know how to type the triangle so I'll say delta x

deltaX = Vf + Vo/2 (t). --- changes to --- delta X = (Vf + Vo/2)(Vf - Vo/a)

Now just solve. Multiply both sides by 2a to get rid of denominator on right side
2a delta X = (Vf + Vo)(Vf - Vo). Vf^2 + VfVo - VfVo - Vo^2
2a delta X = Vf^2 - Vo^2 add Vo^2 to both sides
Vo^2 + 2a delta X = Vf^2 or rewrite Vf^2 = Vo^2 + 2a delta X
• Aren't there five equations, not four?
• Yes there is a fifth equation for initial velocity,(Displacement=vt-0.5at squared). Its not as useful as the other formulas since initial velocity is typically known.
• Where are the videos where Khan explains where each equations comes from?
• what is the difference between v=at and v=x/t?
• the first one tells you what your velocity will be if you start from rest and accelerate at a rate of a for time t
the second one tells you what your average velocity is when you have displacement x during a time t
• I got a question on the test like this: "A chef throws pizza dough in the air and catches it at the same height, the dough travels in the air for 1.0 second. What is the velocity of the pizza dough before it is in the air?" I understand that the known values are the time, acceleration and displacement, and the initial velocity is the one we are trying to find, but why is the final velocity assumed to be unknown? (this is stated in the hint) shouldn't the final velocity be 0 since catching it effectively stops the motion of the dough? I was thinking the final velocity might be referring to the velocity right before impact, but other questions say "coming to a stop" to signal a final velocity of 0, shouldn't it be the same for this question as well?
• In kinematic equations we are trying to understand motion, not just state the obvious fact that after an object is done moving it comes to rest. Final velocity refers to the velocity at the end of the fall. When the dough has been caught, that's not the end of the fall, that's after the end of the fall, and we don't care that the hand made the dough stop falling.

The question is basically asking you how fast do you have to throw something upward to get it to be in the air for 1 second.
• why is it delta x but not delta y because delta x usually means change in time
• We use the letter t to talk about time. x usually stands for position in Physics, even though it’s often graphed on the vertical axis, if that’s what you were thinking. Often, x means position in the x direction, and y means position in the y direction. It would just be confusing to have Δx mean change in time and Δy mean change in position.

Does that help?