# Velocity and distance traveled of a particle

• cdx
In summary: So the max distance is when it changes direction at T = 6, and that is the |v| at T = 2. In summary, the position of a particle moving along the x-axis is given by the equation x = C0 + C1t + C2t^2. The acceleration of the particle is -20 m/s^2 and it passes through the origin at t = -2.0s and t = 6.0s. To find the velocity at t = -2.0s, differentiate the equation twice to get v(t) = -20t + C1. The constant C1 can be found by substituting -2 for t and solving for v(t). To find
cdx

## Homework Statement

(I can't seem to get the super and subscripts to work so I apologize if this looks ugly.) The position of a particle moving along the x-axis is given at time t by: x = C0 + C1t + C2t^2. At time t = 4.0s the acceleration of the object is a = -20 m/s^2 and that the object passes through the origin x = 0 at t = -2.0s and t = 6.0s. Find the velocity of the particle at t = -2.0s and the maximum value of the distance of the particle from the origin measured in the positive x-direction.

## Homework Equations

x = (0.5)at^2 + v0t + x0

## The Attempt at a Solution

I've tried three approaches: differentiating with respect to the x equation above, integrating from the acceleration given, and setting up x = 0 = (t + 2)(t - 6) from the information given.

Differentiating and integrating lead me to the same problem.. how to find the constant C in v(t) = -20t + C.

Setting up the factor-like problem I obtain t^2 -4t - 12. Looking at this equation I'm led to believe that C2 = 1, C1 = -4, and C0 = -12 from the coefficients. Taking the derivative of this, I get v(t) = 2t - 4. Deriving again I get 2 ms^2 which is not the -20 ms^2 provided.

I'm not sure how to start solving this equation correctly. Could someone point me in the right direction? Thanks.

I am a bit too lazy to completely work your problem out but from the information given, the acceleration is constant. So you simply differentiate twice to find that $$2C_{2} = -20$$

edit: just made it a bit prettier

Yes, C2 = -10 in my calculation as well and substituting -10 back into v(t) = 2(C2)t + C1, I get v(t) = -20t + C1. How do I find C1 so that I can substitute -2 for t to solve for v(t)? Or is my answer for the velocity of the particle at time t = -2s simply v(-2) = 40 + C1 m/s? If I don't have a value for C1 then how do I find the maximum value of the distance of the particle in my original x equation?

You know where it is at t = -2 and at t = 6

It must have reversed direction in that 8 seconds. That means at T = 2 it's |V| is 0.

Since a is constant then |v| = a*t = a*4sec at the origin.

## 1. What is the relationship between velocity and distance traveled?

The relationship between velocity and distance traveled is defined by the equation: velocity = distance/time. This means that the velocity of a particle is directly proportional to the distance it travels and inversely proportional to the time it takes to travel that distance.

## 2. How do you calculate the velocity of a particle?

The velocity of a particle can be calculated by dividing the distance traveled by the time it takes to travel that distance. The formula is: velocity = distance/time. Velocity is typically measured in units of distance per time, such as meters per second or kilometers per hour.

## 3. Can the velocity of a particle change over time?

Yes, the velocity of a particle can change over time. This can occur due to changes in the particle's speed, direction, or both. A change in velocity is known as acceleration, and it can be positive (speeding up) or negative (slowing down).

## 4. How does the distance traveled affect the velocity of a particle?

The distance traveled by a particle has a direct effect on its velocity. The longer the distance traveled, the higher the velocity will be. This is because the particle has more time to increase its speed and cover a greater distance.

## 5. Is there a maximum velocity that a particle can reach?

Yes, according to the theory of relativity, there is a maximum velocity that a particle can reach, which is the speed of light. This is approximately 299,792,458 meters per second. At this speed, the particle would have infinite mass and would require an infinite amount of energy to accelerate further.

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