The position of a particle along a straight-line path is defined by t = (t^3 - 6
ID: 1651682 • Letter: T
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