An object is traveling along a linear path according to the equation
s(t) = 4t^3 - 3t^2 + 5 where t is measured in seconds and s(t) measured in meters.
How fast is the object moving at t = 4 seconds?
What is the position of the object when it stops moving?
How far has the object traveled when its acceleration is zero?
Is the object moving towards or away from the origin at t = 3 seconds?
What should I do first to figure this out?
@zxcvbnm1216 this is a pretty standard AP calculus question, so I strongly suggest knowing how to solve these types of problems.
First, you should compute s’(t) and s’’(t). What do they represent? From there, the problems should be pretty straightforward.
I agree with MITer94. Remember the difference between displacement vs distance and velocity vs speed. Displacement and Velocity are vector quantities and depend on the change in final and initial components. Distance and Speed are scalar quantities and depend on the entire path of the object; distance and speed should never be negative values since they do not depend on a point of origin and direction.
The equation you currently have is for distance/displacement. Velocity is the derivative of displacement (position), so take the derivative of that equation to get velocity. Plug t=4 into that to get how fast it is moving. You can also use that to find out the position of the object when it isn’t moving. For the acceleration, take the derivative of the velocity equation.
For general purposes, I’d recommend you re-familiarize yourself with the calculus relationships between position, velocity, and acceleration.
s(t) = 4t^3 - 3t^2 + 5
This is the equation for the object’s position. From here, you figure the equations for velocity and acceleration.
v(t)=ds(t)/dt=(d/dt)(4t^3 - 3t^2 + 5)
a(t)=dv(t)/dt=(d^2 s(t))/dt
Besides knowing this for motion, you should generally recognize that the rate at which something occurs is going to require a derivative. For example, if you get an equation for how many apples johnny has at any time, the rate at which he accumulates the apples at any given time is going to be the derivative of that equation. Also, you should recognize the difference between instantaneous rates of changes (as with the problem you posted and with johnny) and averages rates of change (Δy/Δx).
Let’s take a look at your question:
How fast is the object moving at t = 4 seconds?
-“fast” indicates what quantity? Can this be negative or does it have to be positive?
What is the position of the object when it stops moving?
-This is a little tricky because it requires two steps. Think about the problem backwards: What variable is used in the equation for position? It’s time. So how do you figure out the time at which it stops moving? It’s velocity is going to be zero. Then you go backwards again. Solve for t when v(t)=0, then plug that into s(t).
How far has the object traveled when its acceleration is zero?
-Same thing: Figure the t at which a(t) is zero and plug it into the equation for position.
Is the object moving towards or away from the origin at t = 3 seconds?
-You have to think about this one a little. Basically, you show direction by choosing one direction to be positive and the other to be negative. So -5 for position actually just means 5 in the other direction. It gets a little more complicated to figure it from here. If the object is already in the positive direction, then the velocity must be positive to be moving away from the origin. BUT you also have to remember that if it’s located in the negative direction, then velocity is going to be negative to be moving away from the origin.