the motion of particle along a straight line is described by the equation x=6+4t2 -t4.where x is in meters and t is in seconds. find the position, velocity and acceleration of the particle at t=2seconds...
please help..
Position at t=2 s is obtained by substituting t=2 in the given equation
Thus x= 6 + 4 (2)2 – (2)4 = 6m
Velocity is rate of change of position
v= dx/dt = 8t -4t3 = 8(2) – 4 (2)3 = -16 m/s
acceleration is rate of change of velocity
thus a = dv/dt = 8 – 12 t2 = 8 – 12(2)2= -40 m/s2.