Question (a)
The displacement is equal to
Δ X ⃗ = X f ⃗ − X i ⃗ \Delta \vec{X}=\vec{X_{f}}-\vec{X_{i}} Δ X = X f − X i
Where
Final position x f ⃗ = 27 c m \vec{x_{f}}=27cm x f = 27 c m Initial position x i ⃗ = 45 c m \vec{x_{i}}=45cm x i = 45 c m
Evaluating numerically.
Δ X ⃗ = 27 c m − 45 c m = − 18 c m \Delta \vec{X}=27cm-45cm=-18cm Δ X = 27 c m − 45 c m = − 18 c m
Displacement is
Δ X ⃗ = − 18 c m \boxed{\Delta \vec{X}=-18cm} Δ X = − 18 c m
Question (b)
The speed is equal to:V = d t V=\frac{d}{t} V = t d
Where
Distance traveled (Displacement module) d = ∣ Δ x ⃗ ∣ = 18 c m d=|\Delta \vec{x}|=18cm d = ∣Δ x ∣ = 18 c m The time is: t = ? ? t=?? t = ??
The problem time is not given, but it is if one considers that t = 2s then, the speed is:
V = 18 c m 2 s = 9 c m s V=\frac{18cm}{2s}=9\frac{cm}{s} V = 2 s 18 c m = 9 s c m
Now if you were asked to calculate the velocity then you should use the equation
v ⃗ = Δ x ⃗ t \vec{v}=\frac{\Delta \vec{x}}{t} v = t Δ x
Where
Displacement Δ x ⃗ = − 18 c m \Delta \vec{x}=-18cm Δ x = − 18 c m the time is t = ? ? t=?? t = ??
In this case the sign is maintained, and if it is calculated using t = 2s it has to.
v ⃗ = − 18 c m 2 s = − 9 c m s \vec{v}=\frac{-18cm}{2s}=-9\frac{cm}{s} v = 2 s − 18 c m = − 9 s c m
Note: It takes a time where the displacement occurs to calculate speed and velocity, but here are two examples.
Comments