# 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?

## Answers and Replies

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. :) :) :)