f Facebook c Dwww.webassign.net/web/Student/Assignment-Responses/submit?dep-1350
ID: 2858997 • Letter: F
Question
f Facebook c Dwww.webassign.net/web/Student/Assignment-Responses/submit?dep-13504845 / y A-28a 3.5-3.7 A particle moves according to a law of motion s = f(t), t 0, where t is measured in seconds and s in feet. rt) = 0.01t-006t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.041) _ 0.18t (b) What is the velocity after 5 s? v(5) 0.5 ft/s (c) When is the particle at rest? t=0 t = 912 S (smaller value) s (larger value) (d) When is the particle moving in the positive direction? (Enter your answer using interval notation.) ,00 (e) Find the total distance traveled during the first 11 s. (Round your answer to two decimal places.) 70.71 ft (f) Find the acceleration at time t (in ft/s2) a(t)= 0.12,2-0.181 X m Cortana. Ask me anything 7:05 PM 3/2/2016 DIExplanation / Answer
You have got correct untill part d , so I'm continuing from part e
e) total distance = integral 0 to 11( 0.01t^4-0.06t^3) dt
S = 0.01 *t^5 /5 -0.06 t^4/4 from 0 to 11
S = 0.01 *11^5/5 -0.06 11^4/4 = 102.487 ft -----ANSWER e
f) you have got v(t) = 0.04t^3 -0.18 *t^2
so , acceleration = dv/dt = d/dt( 0.04t^3 -0.18 t^2)
a = 0.04*3 t^2 -0.18*2 t
a = 0.12t^2 -0.36t -------ANSWER f
g) a(5) = 0.12*5^2-0.36*5 = 1.2 -----ANSWER g
for sppeding up we need to have dv/dt =a >0
=> 0.12t^2 -0.36 t>0
0.12 t *(t-3) >0
=> t >0 and t> 3 => t>3 ------ANSWER increasing
decreasing =>
=> 0.12t^2 -0.36 t < 0
=> 0.12 * t *(t-3) <0
=> t-3<0 => t<3 ----ANSWER decreasing
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.