Aparticle moves along a straight line and its position at time t is given by s(t)=2t3−15t2+24t where s is measured in feet and t in seconds. . a) find the velocity (in ft/sec) of the particle at time t=0. b) the particle stops moving twice, once when t=a and again when t=b where a

s(t) = 2 t³ - 24 t² + 72 t 
v(t) = 6 t² - 48 t + 72 

At t = 0, v(0) = 72 ft/s 

v(t) = 0 when 
6 t² - 48 t + 72 = 0 
t² - 8 t + 12 = 0 
(t - 2)(t - 6) = 0 
t = 2 s or t = 6 s 

At t = 16 s 
s(16) = 2 (16)³ - 24 (16)² + 72 (16) = 3200 ft 

The distances traveled by the particle 
* between 0 s and 2 s: d1 = |s(2) - s(0)| = |2*(2)³ - 24*(2)² + 72*(2) - 0| = 64 - 0 = 64 ft ft 
* between 2 s and 6 s: d2 = |s(6) - s(2)| = |2*(6)³ - 24*(6)² + 72*(6) - 64| = |0-64| = 64 ft 
* between 6s and 16s: d3 = |s(16) - s(6)| = |3200 - 0| = 3200 ft 

The TOTAL distance the particle travels between time 0 and time 16: 
d = d1 + d2 + d3 = 64 + 64 + 3200 = 3328 ft

I hope my answer has come to your help. Thank you for posting your question here in .

Liquid. good luck : )

answer and explanation;

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