Q15) Given a particle moving in one dimension with position function x(t) = 5t
3
– 8t
2 + 20:
a) Determine the velocity v as a function of time t. b) Determine the acceleration a as a function of time t.
c) At what time(s) will the particle come to a momentary stop? d) At what time (s) will the particle's
acceleration be zero?
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X (t) = 5t³ - 8t² + 20
a) V (t) = dx / dt
= 15t² - 16t
b) a (t) = d²x / dt²
= 30t - 16
c.) When v = 0
0 = 15t² -16t
16t = 15t²
t = 16/15 = 1.0667
d ) when a = 0
30t - 16 = 0
30t = 16
t = 0.533
a) V (t) = dx / dt
= 15t² - 16t
b) a (t) = d²x / dt²
= 30t - 16
c.) When v = 0
0 = 15t² -16t
16t = 15t²
t = 16/15 = 1.0667
d ) when a = 0
30t - 16 = 0
30t = 16
t = 0.533
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