Position of a particle moving in a straight line is given by: x(t)=t^3-6t^2+9t+3
ID: 1644143 • Letter: P
Question
Position of a particle moving in a straight line is given by: x(t)=t^3-6t^2+9t+3 where position is in meters and time is in seconds.
a) Find an expression for particle's instantaneous velocity vs time.
b) Find an expression for the particle's instantaneous acceleration vs. time
c) Indicate which values of time the particle is moving in the positive direction, in the negative direction and when the particle is stationary.
d) Indicate for which values of time the particle is speeding up, slowing down or neither.
Explanation / Answer
a) v = dx/dt
d(t^n)/dt = n*t^n-1 formula
v = 3t^2-12t+9 m/s
b) a = dv/dt = 6t-12
c) statoionary means v = 0
at t = 1 sec , v = 3-12+9 = 0
at t = 0 to 1 sec v is positve so motion is positive
after 1 sec it may moves in negative direction
d) speeding up means both v and a has same sign
slowing down means both has opposite signs
t = 0 , v = 9 m/s , a = -12 slowing down
t = 1, v = 0 , a = -6 m/s ^2
t = 2, v = -3 m/s, a = 0
t = 3, v = 0, a = 6 m/s^2
t = 4, v = 9 m/s, a = 12 m/s^2 speeding up