Non-constant Acceleration has me stumped.

[philll]

Homework Statement

An object in a frictionless environment accelerates at a rate which increases by 2 m/s2 every second. after 15 seconds, the object reaches is maximum acceleration, and remains constant for a further 30 seconds before all acceleration ceases and it therefore travels at a constant speed for an indefinite period of time. How far had the object traveled after 60 seconds?

u = 0
t = 60

Homework Equations

s = ut + ½ at²

The Attempt at a Solution

I don't even know where to start. I'd like to do it primarily on my own though, so if someone could point me in the right direction, it would be greatly appreciated.

 

[kamerling]

for non-constant acceleration you have

$$v(t) = v(0) + \int_{0}^{t} a(t) dt$$


and $$x(t) = x(0) + \int_{0}^{t} v(t) dt$$

wich you should know how to do if you try this problem.

 

[Moetasim]

I understand your confusion about non-constant acceleration. It can be a tricky concept to grasp, but with some practice and understanding of the equations involved, you can solve this problem.

First, let's break down the information we have been given. We know that the acceleration of the object increases by 2 m/s2 every second, until it reaches its maximum acceleration after 15 seconds. This means that for the first 15 seconds, the object's acceleration is changing. After that, it remains at a constant acceleration for 30 seconds before all acceleration ceases. This means that for the last 30 seconds, the object's acceleration is constant.

To solve this problem, we can use the equation s = ut + ½ at², where s is the distance traveled, u is the initial velocity (which is given as 0), t is the time, and a is the acceleration.

For the first 15 seconds, we can calculate the distance traveled using the equation s = 0(15) + ½ (2)(15)². This gives us a distance of 225 meters.

For the next 30 seconds, we can use the same equation, but now the acceleration is constant at its maximum value. This gives us s = 0(30) + ½ (a)(30)², where a is the maximum acceleration. We can calculate a by using the information given in the problem - the acceleration increases by 2 m/s2 every second, so after 15 seconds it will have increased by 30 m/s2. Therefore, the maximum acceleration is 30 m/s2. Plugging this into the equation, we get s = 0(30) + ½ (30)(30)², which gives us a distance of 1350 meters.

Now, to find the total distance traveled after 60 seconds, we simply add the distances from the first 15 seconds and the next 30 seconds. This gives us a total distance of 225 + 1350 = 1575 meters.

So, after 60 seconds, the object will have traveled a distance of 1575 meters.

I hope this explanation has helped you understand non-constant acceleration a bit better. Keep practicing and you will become more comfortable with it. Good luck on your future problem-solving endeavors!