Question

A particle moves along the x axis. Its position is given by the equation x = 1.5 + 2.9t − 3.6t2 with x in meters and t in seconds. (a) Determine its position when it changes direction. m (b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.) m/s

Answer 1

Given that position is given by the equation

x = 15 + 2.9t - 3.6t^2

Velocity of particle will be given by:

V = dx/dt

V = d[15 + 2.9t - 3.6t^2]/dt

V = 0 + 2.9*1 - 2*3.6*t

V = 2.9 - 7.2*t

Part A.

When velocity will be zero, after that time particle will change it's direction, So

V = 0

2.9 - 7.2*t = 0

t = 2.9/7.2 = 0.403 sec

So at this time position of particle will be

x = 1.5 + 2.9*0.403 - 3.6*0.403^2

x = 2.08 m

Part B.

At t = 0, position of particle is

x = 1.5 + 2.9*0 - 3.6*0^2

x = 1.5 m

Particle will be at this same position when

1.5 = 1.5 + 2.9*t - 3.6*t^2

2.9*t - 3.6*t^2 = 0

t = 2.9/3.6 = 0.805 sec

at t = 0.805 sec, velocity of this particle will be

V = 2.9 - 7.2*0.805

V = -2.90 m/sec

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