Question

The motion of a particle along a straight line described by the equation x=6+4t^2-t^4 Where x is in metres(m) and t is the positive time in seconds(s). Find the position, velocity and acceleration at t=2s. During what time interval is the velocity positive During what time interval is x positive What is the maximum possible velocity attained by the particle

Answer 1

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.

hope it helps u