# The average velocity of the point - numerical value of t

## Homework Statement

During the first t seconds of motion, the average velocity of the point is zero. What is the numerical value of t?

## The Attempt at a Solution

I don't understand this question. I understand that the average velocity is zero if the displacement in the denominator is zero. E.g. if a body is moving in a circle. Or if a body moves to the right and then to the left and distances cancel out. But how could you give the numerical value of t?
I thought t tended to zero (as a limit): 0.0000000001 but it is wrong.
Am I overlooking something obvious?

gneill
Mentor
I think we're missing some context here. Is this part of a larger question? (The problem mentions "the point", not "a point", so it seems to be referring to something previously defined).

• Poetria
Oh yes, silly me. This is an equation from the previous question. I will do it now. That's English for you. Many thanks.

x(t)=2t^2−3t−4

gneill
Mentor
So at time t=0, what's the initial position according to your equation?

When is the position again that value?

• Poetria
It is simple now.
I need the displacement = 0 so

at t = 0 initial position x(i) is -4.

The displacement:
(2t^2−3t−4)-(-4)=0
2t^2−3t=0
t=3/2

I can't assume that t=0 because you can't divide by 0. (the average velocity = displacement/time elapsed)

2*(9/4)-3*(3/2)=9/2-9/2=0 :)

Many thanks. :) :) :)