
Respuesta :
Answer:
a. Negative direction when t = 0s and positive direction when t = 2.83s
b. Negative direction
c. The train would be slowing down when t is between 0 and 1.633 s then speeding up when t is > 1.633 s
Step-by-step explanation:
The velocity function is the derivative of position function
[tex]v(t) = s'(t) = 3t^2 - 8[/tex]
The acceleration function si the derivative of velocity function
[tex]a(t) = v'(t) = 6t[/tex]
a. When s(t) = 0 then
[tex]t^3 - 8t = 0[/tex]
[tex]t(t^2 - 8) = 0[/tex]
t = 0 or
[tex]t^2 - 8 = 0[/tex]
[tex]t = \sqrt{8} = 2.83 s[/tex]
Plug both of the ts into the velocity function and we have
[tex]v(0) = -8 [/tex] so negative direction
[tex]v(\sqrt{8}) = 3*8 - 8 = 16[/tex] so positive direction
b. When a(t) = 0 then 6t = 0 so t = 0. v(0) = -8 so negative direction
c. As a = 6t and t is larger or equal to 0. Then a is also larger or equal to 0. The trains is speeding up if v is positive and slowing down when v is negative
When v is positive
[tex]v(t) > 0 [/tex]
[tex]3t^2 - 8 > 0[/tex]
[tex]t^2 > 8/3 [/tex]
[tex]t > \sqrt{8/3} = 1.633 s[/tex] or [tex]t < -\sqrt{8/3} = -1.633[/tex] (not possible)
Similarly, v is negative when t < 1.633 s
So the train would be slowing down when t is between 0 and 1.633 s then speeding up when t is > 1.633 s
By using the motion equations of the train, we will see that:
a) Β
- v(0) = Β 3*0^2 - 8 = -8 Β (the train goes west)
- v(β8) = 3*β8^2 - 8 = 16 Β (the train goes eastwise)
- v(-β8) = 3*β8^2 - 8 = 16 Β (the train goes eastwise)
b) The direction is west wise.
c)
- (-β, -β8/9) the speed decreases
- (-β8/9, 0) the speed increases (in module, is negative here).
- (0, β8/9) the speed decreases.
- (β8/9, β) the speed increases.
How to work with the motion equations?
We know that the position equation of the train is:
s(t) = t^3 - 8t.
a) To know the direction in which the train moves, we need to find the velocity of the train, which is given by the first differentiation of the position.
v(t) = 3*t^2 - 8
If the velocity is positive, the direction of motion is east wise, if the velocity is negative, the direction of motion is west wise.
The value of t at which s(t) = 0 is:
s(t) = 0 = t^3 - 8*t = t*(t^2 - 8) = t*(t + β8)*(t - β8)
So it is equal to zero for 3 different values of t, which are:
- t = 0
- t = β8
- t = -β8
Evaluating that in the velocity equation we get:
- v(0) = Β 3*0^2 - 8 = -8 Β (the train goes west)
- v(β8) = 3*β8^2 - 8 = 16 Β (the train goes eastwise)
- v(-β8) = 3*β8^2 - 8 = 16 Β (the train goes eastwise)
b) The acceleration is given by the differentiation of the velocity.
a(t) = 2*3*t = 6*t
And the acceleration is 0 only when t = 0, we already know that the velocity is negative when t = 0, which means that the direction is west wise.
c) The velocity is:
v(t) = 3*t^2 - 8
The graph of the parabola can be seen below, there we can see that:
- (-β, -β8/9) the speed decreases
- (-β8/9, 0) the speed increases (in module, is negative here).
- (0, β8/9) the speed decreases.
- (β8/9, β) the speed increases.
The values Β±(β8/9) are the roots of the velocity equation, obtained by solving:
v(t) = 3*t^2 - 8 = 0
Β Β Β Β Β t = Β± β(8/9)
If you want to learn more about equations of motion, you can read:
https://brainly.com/question/605631
