What makes using the velocity vector difficult

Speed on the other hand is defined to be a scalar. So, a velocity vector should be specified using one of the above coordinate systems. Related questions Question b4ef9. What is meant by a component of a vector? How do I find the vertical component of a vector?

How do i find the horizontal component of a vector? Is vector addition commutative? What happens when I multiply a vector by itself?

What is the definition of vector addition? As you turn to the left, your acceleration vector points to the left. We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration. To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed.

Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for non-circular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side.

This is because the car is decelerating as it goes into the curve. The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. In the following, we ignore the effect of air resistance.

This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs. First we need to choose a coordinate system. This is the only force acting on the object.

This force must be equal to the force of gravity at all times, so we therefore know that. Now we use the fact that the acceleration vector is the first derivative of the velocity vector.

Therefore, we can rewrite the last equation in the form. This gives the equation. This gives the position of the object at any time as. How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. Then the position equation becomes.

The height of the archer is Find the horizontal distance the arrow travels before it hits the ground. The equation for the position vector needs to account for the height of the archer in meters. One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level.

In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is. The distance it travels is given by. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets e.

Variations of these laws apply to satellites in orbit around Earth. Thanks to the invariance of the interval and the proper time, every observer agrees on this magnitude.

At first this must seem like a very strange result — every object's velocity 4-vector has the same magnitude, no matter what its world line looks like, and that magnitude is the speed of light?! This is only confusing until one gets accustomed to thinking about velocity 4-vectors differently from velocity 3-vectors. To demonstrate this difference and show this property more clearly, let's express the velocity 4-vector in a specific reference frame:.

The derivatives of the positions with respect to the proper time are not the components of the 3-vector velocity — for that we need derivatives with respect to coordinate time. To this end, we use the chain rule and the relation between proper time and coordinate time time dilation :.

The magnitude-squared of any vector is the dot product of that vector with itself, but for 4-vectors we have to be careful to incorporate the Minkowski metric, so following the matrix multiplication shown in Equation 3. The magnitude of this 4-vector is an invariant, which means that all reference frames will get this same result. Knowing that every object's velocity 4-vector has the same magnitude, and that this magnitude remains the same for all time, may inspire us to ask about acceleration.

On this count, there are two important considerations: First, objects can clearly change the magnitudes of their velocity 3-vectors i. Second, just because the magnitude of a velocity 4-vector doesn't change, it doesn't mean that its direction in spacetime doesn't.

The first is an rotation in space, and the second is a "rotation" between the space and time components. We can construct the acceleration 4-vector in the same manner that we constructed the velocity 4-vector — by taking a derivative with respect to proper time. D erive the 4-vector acceleration components in terms of the 3-vector velocity and 3-vector acceleration for the more general case when these two 3-vectors are not parallel.

The solution will be added after homework 3 is turned-in! There are no fewer than three good ways to solve this. Two of them require clever and powerful relativistic arguments, while the third method consists of brute force algebra.

All three of these are satisfying in their own way. The person who is accelerating knows that they are doing so they can do a test to see that they are not in an inertial frame , so how does the measurement of acceleration made by one observer relate to the measurement made by the other?

At first we might expect both observers to measure the same 3-vector acceleration, but there is a major problem with this. There is no physical law that says that the observer in the accelerated frame can't keep accelerating indefinitely at the same rate — they just need to keep the rocket thrusters set at the same level for as long as they like. So the observer in the inertial frame must witness gradually decreasing acceleration while the accelerated observer observes constant acceleration.

Now use the fact that the magnitude-squared of this 4-vector is an invariant to compare the 3-vector accelerations measured in the two frames.