Question

The motion of a particle in a straight line is defined by the relation x = t⁴ – 12 t² – 40 where x is in meters and t is in sec. Determine the positive x, velocity v and acceleration a of the particle at t = 2 sec.

Answer 1

Dear student

$$\begin{gathered} x=t^4-12t^2-40 \\ puttingt=2 \\ x=2^4-12\times 2^2-40=16-48-40 \\ x=16-88=-72m \\ velocity \\ v=\frac{dx}{dt}=\frac{d(t^4-12t^2-40)}{dt}=4t^3-24t \\ puttingt=2 \\ v=4\times 2^3-24\times 2 \\ v=32-48=-16m/s \\ acceleration \\ a=\frac{dv}{dt}=\frac{d(4t^3-24t)}{dt}=4\times 3t^2-24 \\ a=12t^2-24 \\ a=12\times 2^2-24=48-24=24m/s^2 \\ Regards \end{gathered}$$