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- 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

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