these 3 equations are applicable for constant accelerated motion
each equationand its application is given below
The three equations of motions are:
1)
v = v0 +
aΔ
t 2) x = x0 + v0Δ
t + ½
aΔ
t^2 3)
v^2 = v0
^2 + 2
a(
x −
x0)
These are the equations you learn 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) ; Δ
t = 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 ; Δ
t = 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 ; Δ
t = 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 ; Δ
t = 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.