A particle moves in the xy-plane, starting from the origin at t = 0 with an initial velocity v0 = (20ˆi – 15ˆj) m/s and a constant acceleration of a = 4.0ˆi m/s2. Determine the particle’s speed and direction at t = 5.0 s.
v(t) = v0 + at
so,
v(5) = <20,-15> + 5<ax,ay>
where ax and ay are the components of the acceleration. Since you don't supply them, it's hard to determine v
ax = 4 and ay = 0
To determine the particle's speed and direction at t = 5.0 s, we first need to find the particle's position at this time.
Step 1: Determine the particle's position function:
Since the particle starts from rest at the origin with an initial velocity v0 = (20ˆi – 15ˆj) m/s and a constant acceleration a = 4.0ˆi m/s^2, we can use the equations of motion to find the position function.
The position function in terms of time (t) is given by:
r(t) = r0 + v0t + (1/2)at^2
r0 is the initial position, which is the origin (0, 0) in this case.
v0 is the initial velocity, which is (20ˆi – 15ˆj) m/s.
a is the constant acceleration, which is 4.0ˆi m/s^2.
Plugging in the values, we have:
r(t) = (20ˆi – 15ˆj)t + (1/2)(4.0ˆi)t^2
Step 2: Find the particle's position at t = 5.0 s:
Substitute t = 5.0 s into the position function to find the particle's position at this time.
r(5.0) = (20ˆi – 15ˆj)(5.0) + (1/2)(4.0ˆi)(5.0^2)
= (100ˆi – 75ˆj) + (2.0ˆi)(25.0)
= 100ˆi – 75ˆj + 50ˆi
= 150ˆi – 75ˆj
Therefore, at t = 5.0 s, the particle's position is (150ˆi – 75ˆj).
Step 3: Determine the particle's speed at t = 5.0 s:
The speed of the particle is the magnitude of its velocity vector.
The velocity function in terms of time is given by:
v(t) = v0 + at
Plugging in the values, we have:
v(t) = (20ˆi – 15ˆj) + (4.0ˆi)t
Substituting t = 5.0 s, we can find the velocity at that time.
v(5.0) = (20ˆi – 15ˆj) + (4.0ˆi)(5.0)
= (20 + 20.0)ˆi – 15ˆj
= 40ˆi – 15ˆj
The speed of the particle at t = 5.0 s is the magnitude of the velocity vector:
|v(5.0)| = √[(40)^2 + (-15)^2]
= √[1600 + 225]
= √1825
≈ 42.77 m/s
Therefore, the particle's speed at t = 5.0 s is approximately 42.77 m/s.
Step 4: Determine the particle's direction at t = 5.0 s:
The direction of the particle's velocity vector at t = 5.0 s can be found using the angle θ, where:
θ = arctan(Vy / Vx)
Vx is the x-component of the velocity, which is 40 m/s.
Vy is the y-component of the velocity, which is -15 m/s.
Substituting the values, we have:
θ = arctan((-15) / 40)
= arctan(-0.375)
≈ -20.6°
Therefore, the particle's direction at t = 5.0 s is approximately -20.6°.
To determine the particle's speed and direction at t = 5.0 s, we need to calculate its final velocity at that time.
First, let's calculate the particle's final velocity (vf) using the following equation:
vf = v0 + a * t
Where:
- vf is the final velocity
- v0 is the initial velocity
- a is the acceleration
- t is the time
Given that v0 = (20ˆi – 15ˆj) m/s and a = 4.0ˆi m/s^2, we can substitute these values into the equation to find vf.
vf = (20ˆi – 15ˆj) + (4.0ˆi * 5.0s)
To simplify, we multiply the acceleration vector by the time:
vf = (20ˆi – 15ˆj) + (20ˆi)
Combining like terms, we add the corresponding components:
vf = (20ˆi + 20ˆi) – 15ˆj
Simplifying further, we sum the i-components and subtract the j-component:
vf = 40ˆi - 15ˆj m/s
The final velocity vector vf is given by (40ˆi - 15ˆj) m/s.
To determine the particle's speed and direction, we break down the final velocity vector into its magnitude and direction:
Magnitude of the velocity (speed) = √(vx^2 + vy^2)
Direction of the velocity = arctan(vy / vx)
Where:
- vx is the x-component of the velocity vector
- vy is the y-component of the velocity vector
In this case, the x-component (vx) is 40 m/s and y-component (vy) is -15 m/s.
Let's plug these values into the magnitude and direction formulas:
Magnitude of the velocity (speed) = √((40)^2 + (-15)^2)
Simplifying further:
Magnitude of the velocity (speed) = √(1600 + 225) = √1825 m/s
Direction of the velocity = arctan((-15) / (40))
Simplifying further:
Direction of the velocity = arctan(-0.375) = -20.56 degrees (approximately; rounded to two decimal places)
Therefore, at t = 5.0 s, the particle's speed is approximately √1825 m/s, and its direction is approximately -20.56 degrees relative to the positive x-axis.