Solution
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To calculate the magnitude of acceleration, you need to understand that acceleration is a vector quantity, which means it has both magnitude and direction. The magnitude of acceleration tells us how quickly the velocity of an object is changing. The formula to calculate the magnitude of acceleration depends on the information you have. Here are the common scenarios and the corresponding formulas:
To calculate the magnitude of acceleration, you need to understand that acceleration is a vector quantity, which means it has both magnitude and direction. The magnitude of acceleration tells us how quickly the velocity of an object is changing. The formula to calculate the magnitude of acceleration depends on the information you have. Here are the common scenarios and the corresponding formulas:
1. When initial velocity, final velocity, and time are known:
If you know the initial velocity ($v_i$), final velocity ($v_f$), and the time ($t$) it takes for the velocity to change, you can use the following formula:
where:
- $a$ is the magnitude of acceleration
- $v_f$ is the final velocity
- $v_i$ is the initial velocity
- $t$ is the time over which the change in velocity occurs
2. When distance and velocities are known (using kinematic equations):
If you know the distance ($d$) an object has traveled, its initial velocity ($v_i$), and its final velocity ($v_f$), you can use one of the kinematic equations:
Rearranging the equation to solve for acceleration ($a$) gives:
where:
- $a$ is the magnitude of acceleration
- $v_f$ is the final velocity
- $v_i$ is the initial velocity
- $d$ is the distance
3. When force and mass are known (using Newton's second law):
If you know the net force ($F$) acting on an object and the mass ($m$) of the object, you can use Newton's second law:
Rearranging the equation to solve for acceleration ($a$) gives:
where:
- $a$ is the magnitude of acceleration
- $F$ is the net force
- $m$ is the mass of the object
Let's go through an example using the first formula:
Suppose a car increases its velocity from 10 m/s to 30 m/s in 5 seconds. To find the magnitude of the car's acceleration, we can use the first formula:
Given:
- Initial velocity ($v_i$) = 10 m/s
- Final velocity ($v_f$) = 30 m/s
- Time ($t$) = 5 s
Using the formula:
Substitute the given values:
Calculate the change in velocity:
Now, divide the change in velocity by the time:
So, the magnitude of the car's acceleration is $4 \, \text{m/s}^2$.
1. When initial velocity, final velocity, and time are known:
If you know the initial velocity ($v_i$), final velocity ($v_f$), and the time ($t$) it takes for the velocity to change, you can use the following formula:
where:
- $a$ is the magnitude of acceleration
- $v_f$ is the final velocity
- $v_i$ is the initial velocity
- $t$ is the time over which the change in velocity occurs
2. When distance and velocities are known (using kinematic equations):
If you know the distance ($d$) an object has traveled, its initial velocity ($v_i$), and its final velocity ($v_f$), you can use one of the kinematic equations:
Rearranging the equation to solve for acceleration ($a$) gives:
where:
- $a$ is the magnitude of acceleration
- $v_f$ is the final velocity
- $v_i$ is the initial velocity
- $d$ is the distance
3. When force and mass are known (using Newton's second law):
If you know the net force ($F$) acting on an object and the mass ($m$) of the object, you can use Newton's second law:
Rearranging the equation to solve for acceleration ($a$) gives:
where:
- $a$ is the magnitude of acceleration
- $F$ is the net force
- $m$ is the mass of the object
Let's go through an example using the first formula:
Suppose a car increases its velocity from 10 m/s to 30 m/s in 5 seconds. To find the magnitude of the car's acceleration, we can use the first formula:
Given:
- Initial velocity ($v_i$) = 10 m/s
- Final velocity ($v_f$) = 30 m/s
- Time ($t$) = 5 s
Using the formula:
Substitute the given values:
Calculate the change in velocity:
Now, divide the change in velocity by the time:
So, the magnitude of the car's acceleration is $4 \, \text{m/s}^2$.
Summary
Example part 1 - Normal and Tangential Acceleration Given the Equation of Velocity in Vector Form - Part 1
Exercise 11-An alien spaceship is flying at a trajectory parametrized by r(t)=(sin2t,3,cos2t)
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Let's take a moment to summarize.
In circular motion, we're moving around a circle with a constant radius.
The way we can describe the position is the way we did below
in red xx-hat yy-hat in Cartesian coordinates.
We know that x is r cosine Theta and y is r sine Theta.
The way that we write the r vector,
the position vector is like this.
That's really the most important thing to remember is that this
is the position vector in Cartesian coordinates.
Then we took a derivative of that to find the velocity vector.
If you recall the velocity vector,
we had to use the chain rule to do a derivative over time.
What we ended up with was this whole thing multiplied by Omega^2.
Omega, of course, being the difference in the angle over
the difference in time or the angular velocity.
If you recall, the magnitude of the velocity vector was
equal to Omega r and the angle was tangential,
was orthogonal to the circle itself.
Then we did another derivative to find our acceleration vector.
The first part which ended up being
our radial acceleration is in the direction towards the origin,
towards the center of the circle.
Its magnitude was Omega^2r.
The second part, the tangential acceleration was
in the same tangential direction or the same orthogonal direction.
The tangential acceleration a Theta was in the direction of
the v vector with the value of Alpha r. That's the end of the lecture. Thank you.
This video explains the position, velocity, and acceleration vectors of a particle moving in a circular motion. The position vector is written as r = r cosine Theta and y = r sine Theta. The velocity vector is derived from the position vector and has a magnitude of Omega r and an angle that is orthogonal to the circle. The acceleration vector is composed of two parts: a radial acceleration with a magnitude of Omega^2r and a tangential acceleration with a magnitude of Alpha r.
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