Question

The motion of a particle in a straight line is defined by the relation x = t^4-12t^2_40
where x is in metres and t is in sec. The accleration of the particle at t = 2s is​

Answer 1

Answer:

At t=2sec position x=(2^4-12*2^2-40)=-80 (-ve sign indicates particle moving in -ve x direction)

velocity v=dx/dt=4*t^3-24*t

at t=2sec v=4*2^3-24*2=-16m/sec

acceleration=dv/dt=12*t^2-24

at t=2sec a=12*2^2-24=24m/sec^2