An object moves in the xy-plane with coordinates: x=4cos(20t )(m) and y=4sin(20t )(m).
a/ Find the object’s distance from the origin and the speed of the object at every instant of time. Deduce the characteristics of this motion.
b/ Determine the magnitude and the direction of the acceleration vector at each instant of time.
c/ Sketch the trajectory of this object with the velocity vector and acceleration vector at some instant of time t.
Answers
The distance for origin is 4 m and magnitude of acceleration is 1600 m/s^2.
Explanation:
We are given that:
- X = 4 cos(20 t )(m)
- Y = 4 sin(20 t )(m)
- Vx t = dx / dt = 80 sin ( 20 t )
- Vy t = dy / dt = 80 cos ( 20 t)
- ax (t) = dVx / dt = - 1600 cos ( 20 t)
- ay (t) = dVy / dt = - 1600 sin ( 20 t)
Solution:
(a) Distance from origin = √ x^2 + y^2 = √ ( 16 cos^2 ( 20t ) + sin^2 ( 20 t)
= √ 16 = 4 m
Speed = √ x^2 + y^2 = √ 80^2 ( sin^2 20 t + cos^2 20 t) = 80 m/s
The object is moving in a circle of radius 4 m and is moving in counter clock wise direction.
(b) Magnitude of acceleration = √ ax^2 + ay^2 = 1600 m/s^2
Acceleration is directed toward origin.
Given:
An object moves in the xy-plane with coordinates: x = 4cos(20 t) ( m) and y = 4sin(20 t) (m ) .
To find:
a/ Find the object’s distance from the the origin and the speed of the object at every instant of time. Deduce the characteristics of this motion.
b/ Determine the magnitude and the direction of the acceleration vector at each instant of time.
c/ Sketch the trajectory of this object with the velocity vector and acceleration vector at some instant of time t
Solution:
From given, we have,
x = 4cos(20 t) ( m) and y = 4sin(20 t) (m )
Vx = dx/dt = -80 sin (20t) and Vy = dy/dt = 80 cos (20t)
Ax = dVx/dt = -1600 cos (20t) and Ay = dVy/dt = -1600 sin (20t)
a. The distance from origin,
d = √[x² + y²] = √[16 cos² (20t) + 14 sin² (20t)] = √16 = 4 m
The speed of the object,
Speed = √[Vx² + Vy²] = √[6400 sin² (20t) + 6400 cos² (20t)] = √6400 = 80 m/s
The characteristics of its motion,
The object is moving in a circular path having a radius of 4m moving in counter clockwise direction.
b. The magnitude and the direction of the acceleration
A = √[Ax² + Ay²] = √[(-1600)² cos² (20t) + (-1600)² sin² (20t)] = √1600² = 1600 m/s²
The acceleration is directed towards the origin/centre of the circle.
c. The trajectory of this object with the velocity vector and acceleration vector at some instant of time t is attached below.
