A particle's initial velocity, given as components, is vx = 2.0 m/s and vy = 3.0 m/s. (a) Calculate its velocity 1.5 s later if it experiences an acceleration, given as components, of ax = -1.0 m/s2 and ay = -2.0 m/s2. (b) Calculate the particle's displacement from its initial position at the given time.
I cant figure this out
Just do things the normal way, but using vectors, to keep track of x and y separately.
a(t) = <-1,-2>
v(t) = <2-t,3-2t>
v(1.5) = <0.5,0>
s(t) = <2t - 1/2 t^2, 3t-t^2>
s(1.5) = <1.875,2.25>
No worries, I'll walk you through it step by step.
(a) To calculate the velocity of the particle 1.5 seconds later, we'll use the following kinematic equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Given:
vi = (vx, vy) = (2.0 m/s, 3.0 m/s)
a = (ax, ay) = (-1.0 m/s², -2.0 m/s²)
t = 1.5 s
To find the final velocity, we'll calculate the velocity components separately. Let's start with the x-component:
vxf = vxi + ax * t
Substituting the given values, we get:
vxf = 2.0 m/s + (-1.0 m/s²) * 1.5 s
vxf = 2.0 m/s - 1.5 m/s
vxf = 0.5 m/s
Now let's calculate the y-component:
vyf = vyi + ay * t
Substituting the given values, we get:
vyf = 3.0 m/s + (-2.0 m/s²) * 1.5 s
vyf = 3.0 m/s - 3.0 m/s
vyf = 0.0 m/s
Therefore, the velocity of the particle 1.5 seconds later is:
vf = (vxf, vyf) = (0.5 m/s, 0.0 m/s)
(b) To calculate the displacement of the particle from its initial position, we'll use the following kinematic equation:
Δx = vxi * t + (1/2) * ax * t^2
Δx represents the displacement in the x-direction.
Substituting the given values, we get:
Δx = (2.0 m/s) * 1.5 s + (1/2) * (-1.0 m/s²) * (1.5 s)^2
Δx = 3.0 m + (1/2) * (-1.0 m/s²) * 2.25 s²
Δx = 3.0 m + (-0.5625 m)
Δx = 2.4375 m
Therefore, the particle's displacement from its initial position at the given time is approximately 2.4375 meters.