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.

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