The position function of a particle moving along an x-axis is given by x(t) = 2t

Question

The position function of a particle moving along an x-axis is given by x(t) = 2t^2 - 11t + 12 where x is measured in meters and t in seconds. At what time, if ever, is the particle at the origin x = 0 ? When is the particle's speed equal zero? Is the acceleration ever equal to zero? What is the magnitude of the velocity at 2s? An airplane whose ground speed in still air is 500 m/s, is flying due west. If there is a steady wind at 20 m/s directed towards the south, What is the magnitude of the resultant speed of the airplane? In what direction relative to the x-axis must it now sail to reach its original destination?

Explanation / Answer

5)

a) let time t, the particle is at origin.

use, x(t) = 2*t^2-11*t + 12

0 = 2*t^2-11*t + 12

on solving the above equation we get

t = 1.5 s and t = 4s

b) let time t, v = 0

v = dx/dt

= 2*2*t - 11 + 0

0 = 4*t - 11

==> t = 4/11

= 0.3636 s

c) a = dv/dt

= 4 - 0

= 4

no, acceleration is constant through out the motion.

d) at t = 2s,

v = 4*2 - 11

= -3 m

magnitude of velocity,
|v| = 3 m/s

6)

a) v = sqrt(500^2 + 20^2)

= 500.4 m/s

b) 500*sin(theta) = 20

theta = sin^-1(20/500)

= 2.29 degrees

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