Homework help....please!?

Question

If the positon of an object is x(t)=t + t^3/3 meters then what is the position when the velocity is 0, 1, 2 m/s? What is the position when the acceleration is 0, 1, 2 m/s^2?

Answer 1

x is the position which depends on the time t.

If you differentiate the above equation, you'll get
v(t) = (differentiated equation)
where v = velocity now.
Using the velocity equation now, substitute when v = 0 , what is t
when v=1 what is t, and when v=2 what is t value?

Now you have 3 values of "t" for the various velocities rite? Substitute these t's into the original position equation. and DONE!

Similarly when you differentiate velocity equation, you'll get acceleration equation. Same process..get t's value and sub into positional equation.

I assume you know how to perform your differentiation.

Cheers!


** i must specify don't confuse v(t) as (v x t) . There are different
v(t) = velocity at time t ...
so when you substitute v=0 .. it simply means v(t) = 0
hope you understand.

Answer 2

Velocity is d(x)/dt = t^2; solve for t for each velocity given, and plug each value back into x(t) to get position.

Acceleration is d(v)/dt = 2*t. Again, make this equal to each acceleration given to get t for each value and plug that back into v(x)