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AN630078
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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