The position of a particle moving along an x axis is given by x = 19t2 - 3.0t3,

Question

The position of a particle moving along an x axis is given by x = 19t2 - 3.0t3, where x is in meters and t is in seconds. (a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s. x = 90 Correct: Your answer is correct. m v = 33 Correct: Your answer is correct. m/s a = Incorrect: Your answer is incorrect. m/s2 (b) What is the maximum positive coordinate reached by the particle? m At what time is it reached? s (c) What is the maximum positive velocity reached by the particle? m/s At what time is it reached? s (d) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? m/s2 (e) Determine the average velocity of the particle between t = 0 and t = 3 s. m/s

Explanation / Answer

a) x = 19t^2 - 3t^3
1st Deriv v = 38t - 9t^2
2nd Deriv a = 38 - 18t

Therefore position x = 90m
velocity v = 33m/s
acceleration a = 16m/s^2 (deceleration)

b) maximum positive coordinate reached by the particle is when the velocity is equal to zero to find time. the sub into position equatin to find distance.
time at which velocity equals zero is 4.2222sec.
Distance at which this occurs is 112.9m

c) maximum velocity is achieved when acceleration is equal to zero. Find time from acceleration equation and substitute into velocity equation to find v.

time at which acceleration equals zero is 2.1111sec
Velocity at which this occurs is 40.111m/s

d) The acceleration of the particle at the instant the particle is not moving, ie has zero velocity. Determine time at which zero velocity(other than t=0). This is 4.222sec. Substitute this into acceleration equation.
This gives acceleration at -38m/s^2 (which is 38m/s^2 decelleration)

e)The average velocity between t=0 and t=3 sec is as follows;
distance travelled between this time is 90m.
Considering that this is ondimensional travel it can be calculated as simply as velocity = displacement/time
v = 90/3 = 30m/s

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