The distance travelled by a particle from a fix... - JAMB Mathematics 2009 Question
The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5)cm. Find the minimum distance that the particle can cover from the fixed point.
A
2.3 cm
B
4.0 cm
C
5.2 cm
D
6.0 cm
correct option: c
S = t3 - t2 - t + 5
ds/dt = 3t2 - 2t - 1
As ds = 0
3t2 - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -1/3
At min pt t = -1/3
S = t3 - t2 - t + 5 put t = -1/3
= (-1/3)3 - (-1/3)2 - (-1/3) + 5
= -1/27 - 1/9 + 1/3 + 5
= 140 / 27
= 5.2
ds/dt = 3t2 - 2t - 1
As ds = 0
3t2 - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -1/3
At min pt t = -1/3
S = t3 - t2 - t + 5 put t = -1/3
= (-1/3)3 - (-1/3)2 - (-1/3) + 5
= -1/27 - 1/9 + 1/3 + 5
= 140 / 27
= 5.2
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