Rectilinear motion word question (differintiation)

[jgarcia]

Hi all.

I kindly need help understanding what this question is asking for and solving for it in the right manner.

s(t) is the position of a particle moving along a straight line at time t.

a) Find find the velocity and acceleration of the particle.

b) Find all times in the given interval when the particle is stationary.

s(t) = t² - 2t + 6 for 0 ≤ t ≤ 2

Here is my approach so far:

position s(t) is somewhere between 0 and 2 ∴ V(s(t)) and A(V)

so I first need to find for the change in position by finding the slope, or the derivative of s(t)

s'(t) = t² - 2t + 6
s'(t) = 2t - 2 + 0
s'(t) = 2t - 2
s'(t) = 2 = 2t
s'(t) = 2/2 = 2t/2
s'(t) = 1 ∴ m=1



because t is somewhere between 0 and 2 then the change is -2

y-y₁=m(x-x₁)
y-(-2)=1(x-2)
y=x-2-2
y=x-4



so for a) I have V(t-4) = s/t (speed/time) and A(V(t-4))=ds/t (change in speed/time)

for b) I have that the particle is always moving.

 

[stapel]

s(t) is the position of a particle moving along a straight line at time t.

a) Find find the velocity and acceleration of the particle.

b) Find all times in the given interval when the particle is stationary.

s(t) = t² - 2t + 6 for 0 ≤ t ≤ 2

Here is my approach so far:

position s(t) is somewhere between 0 and 2

No; it is the time t that varies between zero and two. The position is a function of that time t.

∴ V(s(t)) and A(V)

I'm sorry, but I don't understand what you mean here. What is the meaning of the "therefor"? Why do you think that velocity is a function which is evaluated at s(t), rather than being the derivative of s(t)? Why do you think that the acceleration is evaluated at the velocity, rather than being the derivative of the velocity function?

s'(t) = t² - 2t + 6
s'(t) = 2t - 2 + 0
s'(t) = 2t - 2
s'(t) = 2 = 2t

How did you get this last line above? What does it mean?