# Variable acceleration, time, distance

• Markisen
In summary, a particle is moving along a straight line with an acceleration of k(t+2) and initial velocity of 0. The position of the particle as a function of time is x=(kt^3)/6+kt^2+5. To find the time and position when the particle reaches a velocity of 16m/s, the equation 16=(kt^2)/2+2kt is solved for t, which is then inserted into x=(kt^3)/6+kt^2+5 to calculate the distance. The final answer may seem complicated, but it is correct.
Markisen

## Homework Statement

A particle moves along a straight line. At t=0 the acceleration is k(t+2), initial velocity = 0 and x=5m.
a) Write down the position of the particle as a function of time.
b) At what point in time does the particle reach the velocity 16m/s, and where is the particle at this time?

## Homework Equations

$$a=k(t+2)=kt+2k$$
$$v=\int a\textit{dt}=(kt^2)/2+2kt+C_1$$
$$x=\int v\textit{dt}=(kt^3)/6+kt^2+C_2$$

## The Attempt at a Solution

I think I got part a) With the above equations that I calculated and with the initial values inserted I get: $$x=(kt^3)/6+kt^2+5$$

for part b I'm more unsure that I'm correct. Entering the velocity as 16m/s I get the equation: $$16=(kt^2)/2+2kt$$ which when calculated gets the time. $$t=-2+4\sqrt{1+{64/k}^2}$$ and then this time is inserted into $$x=(kt^3)/6+kt^2+5$$ to calculate the distance. This gives me $$1/6\,k \left( -2+4 \,\sqrt {1+64\,{k}^{-2}} \right) ^{3}+k \left( -2+4\,\sqrt {1+64\,{k}^{-2}} \right) ^{2}+5$$ Seems a bit to complicated to be right? Or is it right and I'm just confused

Thanks for the help!

Note that
$$x=\int v\textit{dt}=(kt^3)/6+kt^2+C_1 t + C_2,$$
you forgot to integrate the C1 term.

Finally, you can probably simplify your answer a bit but your way of calculating it looks correct. I am afraid that you are just confused and it is just an ugly answer

Your solution for part a) is correct. However, for part b), your equations are not correct. The time at which the particle reaches a velocity of 16m/s should be when v=16, not when a=16. So, the correct equation to use would be v=kt+2k, and solving for t will give you the time at which the particle reaches a velocity of 16m/s. Then, to find the position of the particle at this time, you can plug in this value of t into your equation for x to get the distance. Your final answer should not involve k, as it is a constant and will cancel out in the calculations. The final answer should be a numerical value for the distance.

## 1. What is variable acceleration?

Variable acceleration refers to a change in the rate at which an object's velocity changes over time. In other words, it is when the speed of an object is changing at a non-constant rate.

## 2. How is variable acceleration calculated?

Variable acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. How does time affect distance in variable acceleration?

In variable acceleration, time affects distance by determining the rate at which an object's velocity changes. The longer the time, the greater the distance an object will travel due to the changes in velocity.

## 4. What is the relationship between acceleration and distance in variable acceleration?

The relationship between acceleration and distance in variable acceleration is directly proportional. This means that as acceleration increases, so does the distance traveled, and vice versa.

## 5. How does variable acceleration impact an object's motion?

Variable acceleration can impact an object's motion by causing it to speed up, slow down, or change direction. This is because the changes in acceleration affect the object's velocity, which in turn affects its position and motion.

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