Answer:
a) positive direction: t < 2s & t>6s ; negative direction: 2s < t < 6s
b) 7 m
c) 71 m
Step-by-step explanation:
Given:
v(t) = 3t^2 -24t +36 [0 , 7]
Find:
a) The value of time when particle is moving in positive direction:
The change in direction of the particle can be determined by v(t) > 0
Hence,
0 < 3t^2 -24t +36
0 < t^2 - 8t + 12
0 < (t - 2)*(t - 6)
t < 2s , t > 6s
The particle travels in positive direction in the interval t < 2s and t > 6s , While it travels in negative direction when 2s < t < 6s.
b) The displacement ds over the given interval [ 0 , 7 ]
ds = integral (v(t)).dt
ds = t^3 -12t^2 +36t
ds = 7^3 -12*7^2 +36*7
ds = 7 m
c) Total distance traveled in the interval:
Total distance= ds(0-2) + ds(2-6) + ds(6-7)
D = 2*(2^3 -12*2^2 +36*2) - 2*(6^3 -12*6^2 +36*6) + 7
D = 2*32 - 2*0 + 7
D = 71 m
