The position of a particle moving along an x axis is given by x = 13.0 t 2 - 3.0

Question

The position of a particle moving along an x axis is given by x = 13.0t2 - 3.00t3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 5.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 5.00 s.

Need the answer as well as the "UNITS" and how you got to the answers.

For part (a) I got a negative number but I don't think you can have a negative postition....Really need help! Thanks.

Explanation / Answer

(a) Taking derivatives of x(t) = 13t^2 %u2013 3t^3 we obtain the velocity and the acceleration functions:
v(t) = 26t %u2013 9t^2 and a(t) = 26 %u2013 18t

with length in meters and time in seconds. Plugging in the value t = 5 yields x(3) =-50 m.

(b) Similarly, plugging in the value t = 5 yields v(3) = -95 m/s.

(c) For t = 5, a(3) = %u201364 m/s2

(d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t =26/9 = 2.888 s for the time of maximum x.

Plugging t =2.888 into the equation for x leads to x =36.76 m

for the largest x value reached by the particle.
(e) From (d), we see that the x reaches its maximum at t =2.888 s.
(f) A maximum v requires a = 0, which occurs when t = 26/18 =1.444 s. This, inserted into the velocity
equation, gives vmax = 26 m/s.
(g) From (f), we see that the maximum of v occurs at t = 26/18 = 1.444 s.
(h) In part (e), the particle was (momentarily) motionless at t = 2.888 s. The acceleration at that time is
readily found to be 26 %u2013 18( 2.888) = %u201325.84 m/s2.
(i) The average velocity,values of x at t = 0 and t =5s are
needed; these are, respectively, x = 0 and x =-50 m (found in part (a)). Thus

Vavg=50-0/5-0=10 m/s

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