A particle has velocities 1m/s ,2m/s, 3√3m/s and 4m/s inclined at an angle of 0•, 60•,150• and 300• respectively to the x-axis find the resultant velocity in magnitude and direction
for each vector, split into components using
x = rcosθ
y = rsinθ
add up the x- and y-parts separately, and convert back to polar form. Most of the stuff cancels out.
I get 1 at 120°
Dgjb
Teach the physics
To find the resultant velocity in magnitude and direction, we first need to find the components of each velocity vector along the x-axis and the y-axis. Once we have these components, we can use vector addition to find the resultant velocity.
Let's break down each velocity vector into its x-component and y-component:
1. Velocity of 1 m/s inclined at 0 degrees:
- x-component: 1 m/s * cos(0°) = 1 m/s
- y-component: 1 m/s * sin(0°) = 0 m/s
2. Velocity of 2 m/s inclined at 60 degrees:
- x-component: 2 m/s * cos(60°) = 1 m/s
- y-component: 2 m/s * sin(60°) = √3 m/s
3. Velocity of 3√3 m/s inclined at 150 degrees:
- x-component: 3√3 m/s * cos(150°) = -3 m/s
- y-component: 3√3 m/s * sin(150°) = √3 m/s
4. Velocity of 4 m/s inclined at 300 degrees:
- x-component: 4 m/s * cos(300°) = 2 m/s
- y-component: 4 m/s * sin(300°) = -2√3 m/s
Now let's add up the x-components and y-components separately:
x-component = 1 m/s + 1 m/s + (-3 m/s) + 2 m/s = 1 m/s
y-component = 0 m/s + √3 m/s + √3 m/s - 2√3 m/s = √3 m/s
To find the magnitude of the resultant velocity, we can use the Pythagorean theorem:
Resultant velocity (V) = √(x-component^2 + y-component^2)
= √(1 m/s^2 + (√3 m/s)^2)
= √(1 m^2/s^2 + 3 m^2/s^2)
= √4 m^2/s^2
= 2 m/s
To find the direction of the resultant velocity, we can use the inverse tangent function:
Direction = tan^(-1)(y-component / x-component)
= tan^(-1)(√3 m/s / 1 m/s)
= 60°
Therefore, the resultant velocity has a magnitude of 2 m/s and is inclined at an angle of 60 degrees to the x-axis.