1.Give examples of a one-dimensional motion where
(a) the particle moving along positive x-direction comes to rest
periodically and moves forward.
(b) the particle moving along positive x-direction comes to rest
periodically and moves backward.
2.Give example of a motion where x > 0, v < 0, a > 0 at a particular
instant.
3. An object falling through a fluid is observed to have acceleration
given by a = g – bv where g = gravitational acceleration and b is
constant. After a long time of release, it is observed to fall with
constant speed. What must be the value of constant speed?
4.A particle executes the motion described by ( ) = (1 − − t );
o x t x e γ t ≥ 0 ,
x0 > 0.
(a) Where does the particle start and with what velocity?
(b) Find maximum and minimum values of x (t), v (t), a (t). Show that
x (t) and a (t) increase with time and v (t) decreases with time.
1.
(a) One-dimensional motion where the particle periodically comes to rest and moves forward along the positive x-direction can be observed in the case of a simple harmonic motion. Examples include a mass oscillating on a spring or a pendulum swinging back and forth. In both cases, the particle's position will periodically reach zero velocity and change direction, moving forward along the positive x-direction.
(b) One-dimensional motion where the particle periodically comes to rest and moves backward along the positive x-direction can be observed in the case of a projectile motion. For example, consider a ball thrown vertically upwards. At the highest point of its trajectory, the ball briefly comes to rest and starts moving backward before falling back down along the positive x-direction.
2. An example of a motion where x > 0, v < 0, and a > 0 at a particular instant can be seen in the case of a decelerating object. Let's imagine a car moving along the positive x-direction, then suddenly applies brakes to slow down. At the instant when the car is still moving forward (x > 0), it has a negative velocity (v < 0) and experiences positive acceleration (a > 0) due to the braking force.
3. In the given scenario, where the object falling through a fluid experiences an acceleration given by a = g - bv, and it falls with constant speed after a long time, the value of the constant speed can be found by equating the acceleration to zero. Since the object falls with constant speed, its acceleration must be zero. Therefore, we have:
g - bv = 0
bv = g
v = g/b
So, the value of the constant speed is v = g/b.
4.
(a) The particle starts at position x = x0 and with an initial velocity v0.
(b) To find the maximum and minimum values of x(t), v(t), and a(t), we can differentiate the position function twice with respect to time.
x(t) = x0(1 - e^(-γt))
Differentiating x(t) with respect to time, we get:
v(t) = γx0e^(-γt)
Differentiating v(t) with respect to time, we get:
a(t) = -γ^2x0e^(-γt)
To find the maximum and minimum values, we need to look at the behavior of these functions. We can observe that as time increases, both x(t) and a(t) approach positive infinity, while v(t) approaches zero. Therefore, the maximum value of x(t) and a(t) is positive infinity, while the minimum value of v(t) is zero.
Also, we can see that x(t) and a(t) both increase with time, as the exponential function e^(-γt) decreases with increasing time. On the other hand, v(t) decreases with time as the exponential function e^(-γt) approaches zero.
Therefore, x(t) and a(t) increase with time, while v(t) decreases with time.
1. (a) An example of a one-dimensional motion where the particle periodically comes to rest and moves forward in the positive x-direction is a simple harmonic motion. This can be observed in systems such as a mass-spring system or a pendulum. In both cases, the particle (mass or bob) oscillates back and forth around a fixed point, reaching maximum displacement in both positive and negative directions and coming to rest at those extreme points before reversing its direction.
To understand the motion, we can analyze the forces acting on the particle and apply Newton's laws of motion. In a mass-spring system, for example, the restoring force is proportional to the displacement from the equilibrium position and acts in the opposite direction. This force causes the particle to accelerate towards the equilibrium position, come to rest momentarily, and then accelerate back in the opposite direction due to the inertia of the mass.
(b) An example of a one-dimensional motion where the particle periodically comes to rest and moves backward in the positive x-direction is the motion of a particle undergoing forced oscillations with damping. In this case, an external force is continuously applied in the opposite direction to the motion, causing the particle to gradually slow down, come to rest, and reverse its direction.
2. An example of a motion where the position (x) is greater than zero (x > 0), velocity (v) is negative (v < 0), and acceleration (a) is positive (a > 0) at a particular instant can be seen in a particle moving in the negative x-direction and experiencing an applied force in the positive x-direction. This situation occurs when the force acting on the particle is greater than the opposing forces (such as friction) and accelerates the particle in the positive x-direction.
For instance, consider a car moving in the negative x-direction with a negative velocity (indicating motion in the opposite direction of the positive x-axis). If the car suddenly accelerates due to pressing the accelerator pedal, the velocity becomes more negative (v < 0), and if the net force applied is greater than the opposing forces (e.g., friction or air resistance), the car will experience a positive acceleration (a > 0) while still moving in the negative x-direction.
3. In the given case, the acceleration of the object falling through a fluid is given by a = g - bv, where g is the gravitational acceleration and b is a constant. After a long time of release, it is observed that the object falls with a constant speed.
When an object reaches terminal velocity, which is a constant speed, the net force acting on it becomes zero. This occurs when the upward force due to fluid resistance (which is proportional to the velocity) becomes equal in magnitude to the downward force of gravity. At this point, the object no longer accelerates and falls with a constant speed.
To find out the value of this constant speed, we equate the gravitational force and the fluid resistance force:
mg = bv
where m is the mass of the object. Rearranging the equation, we get:
v = (mg) / b
Thus, the value of the constant speed is given by (mg) / b.
4. (a) The given equation describes the motion of the particle as x(t) = x0(1 - e^(-γt)), where x0 represents the initial position and γ represents a constant.
To determine where the particle starts, we look at the value of x(t) when t = 0:
x(0) = x0(1 - e^(-γ*0))
x(0) = x0(1 - e^0)
x(0) = x0(1 - 1)
x(0) = 0
Therefore, the particle starts at the origin (x = 0) with zero initial velocity.
(b) To find the maximum and minimum values of x(t), v(t), and a(t), we can differentiate x(t) with respect to time to find v(t), and differentiate v(t) to find a(t):
v(t) = dx(t)/dt = x0γe^(-γt)
a(t) = dv(t)/dt = -x0γ^2e^(-γt)
To find the maximum and minimum values, we can analyze the behavior of x(t), v(t), and a(t) as time progresses. As t approaches infinity, the exponential term e^(-γt) approaches zero, and the maximum and minimum values occur when t = infinity:
Maximum value of x(t) = x0(1 - e^(-γ*∞)) = x0
Minimum value of x(t) = x0(1 - e^(-0)) = 0
Maximum value of v(t) = x0γe^(-γ*∞) = 0
Minimum value of v(t) = x0γe^(-γ*0) = x0γ
Maximum value of a(t) = -x0γ^2e^(-γ*∞) = 0
Minimum value of a(t) = -x0γ^2e^(-γ*0) = -x0γ^2
As we can see, x(t) and a(t) increase with time until they reach their maximum values, while v(t) decreases with time until it reaches its minimum value. This behavior can be further understood by analyzing the derivatives and the exponential decay in the equations.