If the position of a particle is given by x = 30 t - 5 t 3 , where x is in meter
ID: 1734917 • Letter: I
Question
If the position of a particle is given by x = 30t - 5t3, where x is in metersand t is in seconds, answer the following questions.(a) When, if ever, is the particle's velocityzero? (Enter the value of t or 'never'.)
s
(b) When is its acceleration a zero?
s
(c) When is a negative? when 0 < t < 1
whent > 0
whent < 0
(d) When is a positive? when t < 0
whent > 0
when 0< t < 1
(a) When, if ever, is the particle's velocityzero? (Enter the value of t or 'never'.)
s
(b) When is its acceleration a zero?
s
(c) When is a negative? when 0 < t < 1
whent > 0
whent < 0
(d) When is a positive? when t < 0
whent > 0
when 0< t < 1
Explanation / Answer
if a position is given by 30t-5t3, then the derivativewill be the velocity. v(t)=s'(t)=30-15t2 Finding when this is equal to 0, we get -30=-15t2 t2=2 t=±2=1.4 sec To find the acceleration, take the second derivative (thederivative of the velocity) a(t)=v'(t)=s''(t)=-30t Set it equal to 0 to get a(t) = 0 when t = 0 To find when a(t) is positive, you can either plug in differentnumbers, or just see that 0 is the only point where anything mighthappen (it is the only time when acceleration is 0, so it mustremain either positive/negative on either side). Check out when t = -1. a(t) = -30(-1) = 30, which ispositive. Check out when t = 1. a(t) = -30(1) = -30, which isnegative. So the answers to the questions are 1.4 sec 0 sec b a