The position of a particle along a straight-line path is defined by t = (t^3 - 6

Question

The position of a particle along a straight-line path is defined by t = (t^3 - 6t^2 - 15t + 7)ft, where t is in seconds. Determine the total distance traveled when t = 8.6 s. Express your answer to three significant figures and include the appropriate units. What are the particles average velocity at the time given in part A? Express your answer to three significant figures and include the appropriate units. What are the particle's average speed at the time given in part A? Express your answer to three significant figures and include the appropriate units. What are the particle's instantaneous velocity at the time given in part A? Express your answer to three significant figures and include the appropriate units. What are the particle's acceleration at the time given in part A? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

velocity v = ds/dt = 3t^2 - 12 t-15

velocity becomes zero at ,   3t^2 - 12 t-15 =0, taking the positive value of t, t = 5 sec

distance travelled in t = 8.6s,

s(5) = 5^3 -6*5^2-15*5+7 = -93

s(0) = 7

s(8.6) = 8.6^3 - 6*8.6^2 -15*8.6 +7 = 70.296

total distance travelled = [7 - -93] + [70.296- - 93] = 263.3

= 263 m answer

partB] Average velocity = delta s/ delta t = [70.296-7]/8.6

= 7.36 m/s

partC] average speed = distance / time = 263.3/7.36

=35.8 m/s

partD] v = 3*8.6^2 - 12*8.6-15

= 104 m/s

partE] a = 6t-12 = 6*8.6-12

= 39.6 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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