Given displacement equation

$x=c_0(t^2-2)+c(t-2)^2$

Expand the expression

First expand $(t-2)^2$:

$(t-2)^2=t^2-4t+4$

So,

$x=c_0(t^2-2)+c(t^2-4t+4)$

$x=c_0t^2-2c_0+ct^2-4ct+4c$

$x=(c_0+c)t^2-4ct+(4c-2c_0)$

Find velocity

Velocity is

$v=\frac{dx}{dt}$

Therefore,

$v=2(c_0+c)t-4c$

Find acceleration

Acceleration is

$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$

Hence,

$a=2(c_0+c)$

Check the options

• A: acceleration is $2(c+c_0)$ ✅ Correct

• B: acceleration is $2c_0$ ❌ Incomplete

• C: acceleration is $2c$ ❌ Incomplete

• D: initial velocity is $4c$ ❌ Since at $t=0$,

  $v(0)=2(c_0+c)(0)-4c=-4c$

  so initial velocity is $-4c$, not $4c$.

Conclusion

The correct option is:

$\boxed{A}$