A particle moves along a straight line and its position at time t is given by s(

Question

A particle moves along a straight line and its position at time t is given by s(t) = 2t^3 - 21t^2 + 60t| where s te measured in feel and t in seconds Find the velocity (in n/sec) ot the parcle at time t = 0 The panicle stops moving twice. once when t = A and agam when t - B where A lessthan B A is and B is What is the position of the particle at time 14? The TOTAL distance the particle travels between time 0 and time 14 .is The TOTAL distance traveled = |s(A) - s(0)| + |s(0)| + |s(B) - s(A)} + |s(14) - s(B)|

Explanation / Answer

a) velocity , v = ds/dt = 6t2 - 42t +60

At t=0, v = 60 ft/s

b) particle stops moving when v = 0

==> 6t2 - 42t +60 = 0

==> t2 - 7t +10 = 0

==> (t - 2) (t - 5) = 0

t = 2 , 5

Since A<B

Therefore, A = 2 s, B =5 s

c) s(14) = 2*14*14*14 - 21*14*14 + 60*14 = 2,212 ft

d) Total distance covered = | s(2) - s(0) | + | s(5) - s(2) | + | s(14) - s(5) |

s(2) = 2*2*2*2*2 - 21*2*2 + 60*2 = 52

s(0) = 0

s(5) = 2*5*5*5 - 21*5*5 + 60*5 = 25

s(14) = 2*14*14*14 - 21*14*14 + 60*14 = 2212

Total distance = | 52 - 0| + | 25 - 52| + | 2212 - 25 |

= 52 +27 + 2187

= 2266 ft

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