Constant Acceleration Equations- PhysPrimer

An intro to the constant acceleration equations, testing them with a motion sensor, and some examples of their use in problem solving.
0:00 Introduction
1:24 Three Constant Acceleration Equations
2:49 Testing Patterns with a Motion Sensor
7:37 Taking Data for Cart Rolling Down Ramp Example
10:00 Diagram for Cart Rolling Down Ramp Example
11:00 An INCORRECT Approach for Cart Example
12:40 CORRECT Solution to Cart Example
16:49 Another Solution to Cart Example
18:24 Projectile Motion
20:59 Falling Object Example
25:15 Try it yourself examples (solutions at end of description)
Opening Image Credit: NASA, ESA, CSA, STScI, Klaus Pontoppidan (STScI), Image Processing: Alyssa Pagan (STScI)
science.nasa.gov/missions/jam...
Solutions to "Try it yourself" examples (and I do recommend trying these yourself BEFORE looking at solution):
1) Rearrange the first equation to solve for time:
t = (v - v0)/a
Substitute this expression for time into the second equation:
x = x0 + v0*(v - v0)/a + 1/2 * a * (v^2 - 2*v*v0 + v0^2)/a^2
Subtract x0 from both sides and multiply both sides by a:
a*(x - x0) = v0*v - v0^2 + 1/2*v^2 - v0*v + 1/2*v0^2
Simplify
a*(x - x0) = 1/2*v^2 - 1/2*v0^2
Multiply both sides by 2 and then add v0^2 to both sides:
v0^2 + 2*a*(x - x0) = v^2
This is the third equation.
2) Recommend drawing a diagram (always) and let's pick up as the positive direction. We have v0 = 8.5m/s, and since it is a projectile, a = -9.8m/s^2. At the top of the motion the vertical velocity should momentarily go to zero (v = 0). So we can use equation (1) to find the time to the peak (0.867s), and then plug that time into equation (2) to find the maximum height (3.69m). You can also directly find the max height by using equation (3) without needing to first find the time (still get 3.69m).
3) DRAW THE DIAGRAM, and let's pick up as the positive direction, and the astronaut's hand as position zero. So, in the motion described, x0 = 0m and x = 0m (it starts and ends at her hands). They throw at 7.2m/s (so this is v0) and we can use the second constant acceleration equation,
x = x0 + v0*t + 1/2*a*t^2
We aren't on the Earth so we can't assume that a=-9.8m/s^2, but we know the time from throw to catch is 4.3s. So,
0 = 0 + (7.2m/s)*(4.3s) + 1/2*a*(4.3s)^2
0 = 30.96m + (9.245s^2)*a
Rearrange
a = -3.35m/s^2

[@richardpepsistein-lr4ss]

understanding physics ain't just for engineers it's real life

[@PhysicistMichael]

And I try to emphasize the critical thinking and analysis skills that are hopefully applicable everywhere

[@NalitaQubit]

Thank you for your hard work. Do you provide private tutoring to adults who love Physics by any chance ? I’m really impressed by your simple ways to explain complex phenomena.