# Kinematics Motion with Variable Acceleration

• AN630078
The velocity can be found by integrating this equation: $$\int_{0}^t v(t)dx = v(t_1) \int_{t_1}^t v(t_2)dx = v(t_3)$$So, in this case, ## v(t) = c_1 x^2 + c_2 x + c_3 ##.

#### AN630078

Homework Statement
Hello, I have been really trying to improve my understanding of mechanics problems which I struggle with frequently. While doing so I came across the question below and have tried to solve it to the best of my ability. However, I am entirely uncertain whether this would be the correct approach of whether I have reached a credible solution. Could anyone advise me upon whether I could improve upon my workings shown here or offer any guidance for similar problems?

A particle of velocity v(t)=4-3t^2 ms^1 is at the Origin at the start of motion. Find the particle's acceleration when it is next at the Origin.
Relevant Equations
s= ∫v dt
a=dv/dt=d^2s/dt^2
The displacement of the particle is;
s= ∫v dt
s= ∫4-3t^2 dt
s=4t-t^3+c
When the particle is at the Origin, t=0;
0=4(0)-(0)^3+c
c=0
So this becomes;
s=4t-t^3
The particle next passes through the origin when;
4t-t^3=0
Factor out the common term -t;
-t(t^2-4)=0
Rewrite t^2-4 as t^2-2^2 and factor by applying the differnce of two squares;
t^2-2^2=(t+2)(t-2)
-t(t+2)(t-2)=0
t=0, t=-2 and t=2

Therefore, the particle must next pass through the Origin when t=2
The acceleration can be found by differentiating velocity with respect to time;
a=dv/dt=d^2s/dt^2
a=0-6t=-6t
When t=2
a=-6(2)=-12ms^-2

Would this be correct? I would be very grateful of any response  AN630078 said:
a=dv/dt=d^2s/dt^2
a=0-6t=-6t
When t=2
a=-6(2)=-12ms^-2

Would this be correct? I would be very grateful of any response  Looks correct to me.

To answer the earlier part of the post regarding the method, there is nothing to improve here. The method is correct and I cannot see any better ways of solving this problem. Importantly, you didn't forget to think about the integration constant (even though it is 0 here) which will probably catch people out in some problems.

Which concepts are you struggling with at the moment?

In terms of other problems of this sort, the only other trick I have seen is where they ask a similar question (i.e. find the velocity and the acceleration at some time ## t ##) and they give ## t = f(x) ##. You might want to think how you could get to the velocity in that sort of situation (Hint: there are two ways, one of which will likely involve much more work) and then how you could obtain an expression for the acceleration. In some problems, sketching a diagram might be of some help.

• AN630078
Master1022 said:
To answer the earlier part of the post regarding the method, there is nothing to improve here. The method is correct and I cannot see any better ways of solving this problem. Importantly, you didn't forget to think about the integration constant (even though it is 0 here) which will probably catch people out in some problems.

Which concepts are you struggling with at the moment?

In terms of other problems of this sort, the only other trick I have seen is where they ask a similar question (i.e. find the velocity and the acceleration at some time ## t ##) and they give ## t = f(x) ##. You might want to think how you could get to the velocity in that sort of situation (Hint: there are two ways, one of which will likely involve much more work) and then how you could obtain an expression for the acceleration. In some problems, sketching a diagram might be of some help.
Thank you very much for your reply. I usually just struggle with deciphering the wording of these questions, when it is sometimes a little more ambiguous than it is here. I am believe that I am so concerned about being unable to solve the problem suitably that I throw myself off anyhow.

Oh, I have not seen a problem like you suggest and admittedly I have been thinking about it but I am uncertain. How would one find the velocity when t=f(x)?

AN630078 said:
Thank you very much for your reply. I usually just struggle with deciphering the wording of these questions, when it is sometimes a little more ambiguous than it is here. I am believe that I am so concerned about being unable to solve the problem suitably that I throw myself off anyhow.
No problem, happy to be of some help. With this line of problems, getting used to interpreting displacement/velocity/acceleration-time graphs will help. Without an example, it is hard to give specific advice about the wording.

As for gaining confidence about problem solving, I think practice will definitely help. As you progress through further education, I have found that the problems require many more steps and can therefore seem daunting (also with the opportunity cost of lost time). However, making mistakes often helps to gain a greater understanding the real concept so mistakes shouldn't always have a negative connotation. Despite the different wordings, many of these problems will be solved by reaching the same conceptual points in the problem. For example, in the problem above, you knew that you needed to find ## s(t) ## and find the ## t_1 ## such that ## s(t_1) = 0 ## and then evaluate ## a(t_1) ##. It is worth thinking about the overall process each time you attempt a question and consider it step-by-step.

AN630078 said:
Oh, I have not seen a problem like you suggest and admittedly I have been thinking about it but I am uncertain. How would one find the velocity when t=f(x)?
I apologize as this slightly veers off the original post. For example, if you were given:
$$t = c_1 x^2 + c_2 x + c_3$$ and you were asked to find ## v(t) ## and ## a(t) ##
We know that ## v = \frac{dx}{dt} ## and therefore we want to work towards ## \frac{dx}{dt} ## (can think about acceleration afterwards). Some may consider this more of a calculus problem rather than a mechanics problem. One immediate answer might be to find x as a function of t. While that is theoretically correct and will work, it looks like a work-intensive method. Is this a faster way of getting to the required derivative?

Perhaps we might want to use the fact that ## \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} ## or use implicit differentiation to get to the result

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