If an object is initially moving with a constant velocity of 20m/s towards positive x-axis and after 5 sec it changes its direction and moving at an angle of 30o with x-axis, find magnitude and direction of its acceleration.
Vo = 20 m/s[0o]
V = 20 m/s[30o]
a = (V-Vo)/t = (20[30o]-20[0o])/5 =
(20*cos30+i20*sin30-20)/5 =
(17.32+10i-20)/5 = (-2.68+10i)/5 =
-0.536 + 2i,(Q2).
Tan Ar = Y/X = 2/-0.536 = -3.73134
Ar = -75o = Reference angle.
A = -75 + 180 = 105o,CCW = 15o W. of N.
a = 2/sin105 = 2.07 m/s^2 @ 105o, CCW.
To find the magnitude and direction of acceleration, we first need to determine the change in velocity.
Given:
Initial velocity, u = 20 m/s
Time, t = 5 sec
The change in velocity can be calculated using the equation:
Change in velocity = Final velocity - Initial velocity
Since the object changes its direction after 5 seconds, the final velocity has a different magnitude and a direction.
To calculate the magnitude of the final velocity, we can use the given angle and the initial magnitude of velocity.
Magnitude of final velocity, v = initial magnitude of velocity = 20 m/s
To find the components of the final velocity in the x and y directions, we can use trigonometry.
The x-component of the final velocity can be calculated as:
v_x = v * cos(theta)
= 20 m/s * cos(30°)
= 20 m/s * √3/2
= 10√3 m/s
The y-component of the final velocity can be calculated as:
v_y = v * sin(theta)
= 20 m/s * sin(30°)
= 20 m/s * 1/2
= 10 m/s
Since the object changes its direction and its x-component of velocity changes sign, the x-component of velocity after 5 seconds will be -10√3 m/s.
Now, we can calculate the change in velocity, Δv, by subtracting the initial velocity from the final velocity:
Δv = v - u = (-10√3 m/s - 20 m/s)i + (10 m/s)j
Once we have the change in velocity, we can find the acceleration using the equation:
Acceleration = Change in velocity / Time
Acceleration = Δv / t = ((-10√3 m/s - 20 m/s)i + (10 m/s)j) / 5 s
To find the magnitude and direction of the acceleration, we can express it in terms of its magnitude and angle with the x-axis.
Magnitude of acceleration, A = √(Ax^2 + Ay^2)
Direction of acceleration, α = tan^(-1)(Ay / Ax)
Plugging in the values, we get:
Magnitude of acceleration, A = √( (-10√3 / 5)^2 + (10 / 5)^2 )
= √(30 + 4)
= √34 m/s^2
Direction of acceleration, α = tan^(-1)( (10 / 5) / (-10√3 / 5) )
= tan^(-1)( -1 / √3 )
= -π/3 radians or -60°
Therefore, the magnitude of the acceleration is √34 m/s^2, and its direction is -60° with the x-axis.