A particle has an initial velocity of 9m/s due east and an acceleration of 2 m/s^2 due west. Find the distance travelled in 5th second?


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Oddman answered
If we take due east to be the positive direction, we can write the equation for the position (x) of the particle at time (t) as
  x = 9t - (1/2)(2)t^2 = t(9-t)
The 5th second is the one ending at t=5. A graph of the travel in that time is seen here.

When we put t=4 and t=5 into the equation above, we find that the value
  x(4) = x(5) = 20 meters
Thus, there is no net travel by the particle in that time period. However, we know the particle is not standing still. The farthest east it will travel is at time t=4.5, at which point
  x(4.5) = 20.25 meters

In the 5th second, the particle travels from 20 m east to 20.25 m east and back again, for a total distance traveled of 0.50 meters. As stated above, its net displacement in that period is zero.

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