Confused about test answer

Question

On my test, there was a question that was ''find the average velocity for the time interval [0,3], given the displacement function of a particle moving in a straight line relative to a point O is s(t) = 5t - t² where 0 ≤ t ≤ 5, where t is in seconds and s is in meters,'' and I got it wrong. I got the answer -2m^-1 by finding the first finding the velocity function by differentiating 5t - t² which yields - 2t + 5. Next, I did s'(2) and s'(5) to find the initial and final velocities. I then found the average velocity via the gradient formula (y₂ - y₁) / (x₂ - x₁) yielding -2ms^-1. Though the answer was 1 meter per second. Am I missing something here or is the answer wrong...?

Answer 1

Dear Sam,

I see you are having some trouble. I hope I can help.

One issue is we need to define what the average velocity is. The average velocity is typically defined by the number of meters displaced (I use this word because the object could have come back). So the typical way you would find this average velocity is to determine where the object is at the two time intervals and divided by the time. So, an equation would look something like V_avg = (s(t_2) - s(t_1))/(t_2-t_1) where t_2 is the final time and t_1 is the inital time.

So once we understand this it is simple to plug in the numbers. For your example, t_1 = 0 and t_2 = 3.

The object at t_1 is s(0) = 5(0) - 0^2 = 0 meters from O.

The object at t_2 is s(3) = 5(3) - 3^2 = 4 meters from O.

Now that we know this, we should be able to determine the average velocity.

V_avg = (4-0)/(3-0) = 4/3 or 1.333 m/s

I understand this is different to what you say the answer should be of 1. I think this may be a typo in the question as the time interval [0,4] would have an average velocity of 1 m/s.

In addition, I would also like to just add please be careful with your definitions. You took the derivative to find the velocity function. If you evaluate this function at a certain time (say for instance at 1 second), this gives you the instantanous velocity at that very moment in time. Then if you take the gradient (or slope) of the velocities you are finding the accelaration.

Hope that helped.

Tim