Non-Constant Acceleration Problem

[Jewelz]

Homework Statement

A new Tesla is designed that can perform a non-constant acceleration for 10 seconds of motion. The magnitude of the acceleration is given as a(t) = 1 m/s⁴t²

Starting from rest, how far does the car travel over this 10 second interval?

Homework Equations

This is what is making the question difficult for me. I am unsure what equations to use with a non-constant acceleration.

The Attempt at a Solution

I tried solving it manipulating the constant velocity and acceleration equations, but everything I have tried has been wrong. Even if someone could point me in the right directions equations wise, I'm sure that would help me a lot.

Thanks

 

[Ssnow]

Hi, you can find the velocity function respect to $t$ that is $v(t)=\int_{0}^{t} a(s)ds$ and after the space $x(t)=\int_{0}^{t}v(s)ds$, put $t=10 s$ in $x(t)$ ...
Ssnow

 

[Ray Vickson]

Homework Statement

A new Tesla is designed that can perform a non-constant acceleration for 10 seconds of motion. The magnitude of the acceleration is given as a(t) = 1 m/s⁴t²

Starting from rest, how far does the car travel over this 10 second interval?

Homework Equations

This is what is making the question difficult for me. I am unsure what equations to use with a non-constant acceleration.

The Attempt at a Solution

I tried solving it manipulating the constant velocity and acceleration equations, but everything I have tried has been wrong. Even if someone could point me in the right directions equations wise, I'm sure that would help me a lot.

Thanks

Your input is hard to read; I assume you mean $a = k t^2,$ where $k = 1 m/s^4.$

Anyway, you get velocity $v$ by integrating $a$ with respect to $t$, and then you get position by integrating $v$. No amount of manipulation of the constant-acceleration formulas can do what you need.

 

[Jewelz]

Your input is hard to read; I assume you mean $a = k t^2,$ where $k = 1 m/s^4.$

Anyway, you get velocity $v$ by integrating $a$ with respect to $t$, and then you get position by integrating $v$. No amount of manipulation of the constant-acceleration formulas can do what you need.

Your assumption is correct.

After integrating the acceleration function with respect to time, with the bounds of the integral from $0$ to $t$, I obtained the function $t^3/3$ for the velocity. Integrating that, from 0 to t for the integral, I got $t^4/12$, and plugging in $t$ (10 seconds), I obtained a final answer of 833.3m traveled.

Does this all sound correct?

 

[PeroK]

Your assumption is correct.

After integrating the acceleration function with respect to time, with the bounds of the integral from $0$ to $t$, I obtained the function $t^3/3$ for the velocity. Integrating that, from 0 to t for the integral, I got $t^4/12$, and plugging in $t$ (10 seconds), I obtained a final answer of 833.3m traveled.

Does this all sound correct?

It does.

 

[Ray Vickson]

Your assumption is correct.

After integrating the acceleration function with respect to time, with the bounds of the integral from $0$ to $t$, I obtained the function $t^3/3$ for the velocity. Integrating that, from 0 to t for the integral, I got $t^4/12$, and plugging in $t$ (10 seconds), I obtained a final answer of 833.3m traveled.

Does this all sound correct?

Yes, perfect.