Answer:
162
Explanation:
In this problem, we know that the position of the particle is described by the equation
[tex]x(t)=2t^4-3t^3[/tex]
where t is the time.
The velocity of the particle can be calculated as the derivate of the position; so we find:
[tex]v(t)=x'(t)=\frac{d}{dt}(2t^4-3t^3)=4\cdot 2t^3 -3\cdot 3t^2=8t^3-9t^2[/tex]
Similarly, the acceleration of the particle can be found by calculating the derivative of the velocity.
By doing so, we find an expression for the acceleration:
[tex]a(t)=v'(t)=\frac{d}{dt}(8t^3-9t^2)=3\cdot 8t^2-2\cdot 9t=24t^2-18t[/tex]
Therefore, this is the expression for the acceleration of the particle.
Now we want to find the acceleration after 3 s. To do so, we substitute the value of t = 3 into the expression above; so we find:
[tex]a(3)=24(3)^2-18(3)=162[/tex]
