5· Question Details A particle moves according to a law of motion sf(t), t 2 0,

Question

5· Question Details A particle moves according to a law of motion sf(t), t 2 0, where t is measured in seconds and s in feet. (a) Find the velocity at time t (in ft/s). v(t) = (b) What is the velocity after 6 ? (Round your answer to two decimal places.) v(6) = ft/s (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (Enter your answer using interval notation.) (e) Find the total distance traveled during the first 12 s. (Round your answer to two decimal places.) ft (f) Find the acceleration at time t (in ft/s2). a(t) = Find the acceleration after 6 s. (Round your answer to three decimal places.) a(6) ft/s2

Explanation / Answer

5)

position s=f(t)=te-t/5

(a)

velocity ,v(t)=f '(t)

v(t)=(1*e-t/5)+(te-t/5(-1/5))

v(t)=(1-(t/5))e-t/5

b)

v(6)=(1-(6/5))e-6/5

v(6)=-(1/5)e-6/5

v(6)= -0.06 ft/s

(c)

at rest, v(t)=0

=>(1-(t/5))e-t/5=0

=>(1-(t/5))=0

=> t =5 s

(d)

positive direction =>v(t)>0

=>(1-(t/5))e-t/5>0

=>(1-(t/5))>0

=>(t/5)<>1

=>t<5

positive direction [0,5)

e)

distance travelled = [f(5)-f(0)]+[f(5)-f(12)]

distance travelled = [5e-5/5-0e-0/5]+[5e-5/5-12e-12/5]

distance travelled = 2.59 ft

(f)

a(t)=v'(t)

a(t)=[(0-(1/5))e-t/5]+[(1-(t/5))e-t/5(-1/5)]

a(t)=[-(1/5)e-t/5]+[(1-(t/5))(-1/5)e-t/5]

a(t)=-(1/5)e-t/5(1+1-(t/5))

a(t)=-(1/5)e-t/5(2-(t/5))

a(t)=(1/25)(t-10)e-t/5

a(6)=(1/25)(6-10)e-6/5

a(6)= -0.05 ft/s2

please rate if helpful. please comment if you have any doubt

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.