Question

pls help me!!! and answer if u actually know!!!

Answer 1

Answer:

it's not clear please send clear pic of question it is difficult to understand it .

Answer 2

Given : S(t) = 3t + 4t² - t³

0 ≤ t ≤ 5

To Find: Average velocity for 0 ≤ t ≤ 4

general expression for velocity , velocity at t = 2

Time between 0 ≤ t ≤ 5 where particle changes direction

Solution:

S(t) = 3t + 4t² - t³

Average velocity for 0 ≤ t ≤ 4

S(0) = 0

S(4) = 3(4) + 4(4)² - 4³  = 12

Average velocity = (12 - 0)/(4 - 0)  = 3 m/s

v(t) = S'(t)  = dS/dt  = 3  + 8t  - 3t²

general expression for velocity v(t) = 3  + 8t  - 3t²

velocity at t = 2   v(2) = 3 + 8(2) - 3(2)²  = 7  m/s

As velocity is positive hence particle moving to its right

3  + 8t  - 3t²

=  3 + 9t - t - 3t²

=  3(1 + 3t) - t(1 + 3t)

= (3 - t)(1 + 3t)

Hence Velocity is positive between  ( -1/3 and 3)

and negative before t < -1/3  and  t > 3

Hence between 0 ≤ t ≤ 5  , particle changes direction at  t = 3

v(t) = 3  + 8t  - 3t²

a(t) = v'(t) = dv/dt  = 8  - 6t

at t = 4   ,  -16  Hence Velocity is decreasing

But magnitude is increasing   ( see the graph)

Learn More:

The displacement s of a particle at a time t is given by s = t³ - 4t² - 5t ...

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