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Question

When we used derivation of motion in numericals

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Solution

The three equations of motions are:

1) v = v0 + aΔt
2) x = x0 + v0Δt + ½aΔt^2
3) v^2 = v0^2 + 2a(xx0)

These are the equations you learn in the physics class at school to calculate the velocity or end position of an object given an acceleration and a time or a distance.

Let me tell you a little secret: there are no 3 equations of motion. The three things you see above follow directly from the definition of a constant acceleration in one direction. But you need a bit of calculus to see that. The 3 above are an easy way to calculate things without needing calculus.

To know which one you need requires a close examination of your physics problem. E.g.
  • Equation 1) you use when the physics teacher asks you to calculate a speed, given some actions during a period of time.
  • Equation 2) you use when the physics teacher asks you to calculate a position, given some actions during a period of time.
  • Equation 3) you use when the physics teachers asks you to calculate a speed, given some actions over some distance.

Lets try this out!

Equation 1: gives you a speed as answer and it needs a time as input.
  • What is the speed of Jan when she travels at 10 m/s without accelerating for 10 seconds?

Answer:

v0 = 10 m/s ; a = 0 m/s^2 (she isn't accelerating) ; Δt = 10 s

So, v = 10 + 0 * 10 = 10 m/s
  • What is the speed of Jan when she travels at 10/ms and decelerates at 10 m/s^2 for 2 seconds?

Answer:

v0 = 10 m/s ; a = -10 m/s^2 ; Δt = 2 s

So, v = 10 + -10 * 2 = - 10 m/s (she travels in the opposite direction)

Equation 2: gives you a position as answer and needs a time as input.
  • Jan travels at 10 m/s without acceleration where is she 10 seconds later?

Answer:

x0 = A (some point in space) ; v0 = 10 m/s ; a = 0 ; Δt = 10 s

So, x = A + 10 * 10 + ½ * 0 * 2^ 2 = A + 100 meter.

I don't know where A is but Jan is 100 meters further on her route.
  • Jan travels at 10 m/s , decelerates at 10 m/s^s for 2 seconds, where is she now?

Answer:

x0 = A (some point in space) ; v0 = 10 m/s ; a = -10 m/s^2 ; Δt = 2 s

So, x = A + 10 * 2 + ½ * - 10 * 2 ^ 2 = A + 20 - 20 = A

I don't know where A is but that is where Jan is now.

Equation 3: gives you a speed given a distance.
  • Jan travels at 10 m/s , from the last problem we know she decelerated at 10 m/s^2 and is again at A. What is her speed?

Answer:

v0 = 10 ; a = -10 m/s^2 ; x - x0 = 0 (she moved right turned and is back in A).

v ^ 2 = 10 ^ 2 + 2 * -10 * 0 = 100 ; | v | = 10 m/s^2 We know that the magnitude of the speed is 10 m/s, but because equation 3 uses squares we don't know the direction.

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