The position of a particle moving along the x axis with a constant acceleration

Question

The position of a particle moving along the x axis with a constant acceleration may be determined from the expression x(t) = bt + ct^2, where b = 4.50 m/s, c = 3.90 m/s^2, and x will be in meters when t is in seconds. Determine the following. (a) average velocity of the particle over the time interval t = (1 rightarrow 3) s (Indicate the direction with the sign or your answer.) m/s (b) instantaneous velocity of the particle at t = 2 s, the midpoint of the above time interval (Indicate the direction with the sign of your answer.) m/s (c) Which of the following statements is the best explanation for the relationship between your answers to parts (a) and (b)? The answers are different because the acceleration is in the opposite direction of the velocity. The answers are the same because the acceleration is zero. The answers are different because the acceleration is in the same direction as the velocity. The answers are the same because the acceleration is in the same direction as the velocity. The answers are the same because the acceleration is constant.

Explanation / Answer

given
position x(t) = bt + ct^2
b = 4.5 m/s
c = 3.9 m/s/s

a. Vav from T = 1 TO T = 3
   Vav = displacement/time
   Vav = [(x3) - (x1)]/(3 - 1) = [3.9(9 - 1) + 4.5(3 - 1)]/2 = 20.1 m/s
b. at t= 2 s
   dx(t)/dt = instantaneous velocity = b + 2ct = 4.5 + 2*3.9*2 = 20.1 m/s
c. the answers are same because the acceleration is constant
   because
   acceleration = d^2(x(t))/dt^2 = 2c -> constant
   also, for constant acceleration
   v(t = 2) = v(t = 0) + 2c*2 = 5c
   v(t = 1) = v(t = 0) + 2c*1 = 3c
   v(t = 3) = v(t = 0) + 2c*3 = 7c
   vav = (v(t = 3)+v(t = 1))/2 = 5c = v(t = 2)

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