# Position of a particle

## Homework Statement

The position of a particle moving along the x axis varies in time according to the expression x = 6t2, where x is in meters and t is in seconds. Evaluate its position at the following times.
(a) t=2.5s
(b) t=2.5s + Δt
(c) Evaluate the limit of Δxt as Δt approaches zero to find the velocity at t = 2.50 s.

Δxt

## The Attempt at a Solution

(a) was simple, just plug in 2.5s and get 37.5m
(b) the problem doesn't explain well enough, but i finally got it, plug in 2.5 + Δt for t and I got 6(2.5 + Δt)^2
(c) I don't even know how to do this, it's been 4 years since I've taken classes of higher math and I must have forgotten most. The answer provided is 30 but I don't know the steps to get there.

Related Introductory Physics Homework Help News on Phys.org
SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

The position of a particle moving along the x axis varies in time according to the expression x = 6t2, where x is in meters and t is in seconds. Evaluate its position at the following times.
(a) t=2.5s
(b) t=2.5s + Δt
(c) Evaluate the limit of Δxt as Δt approaches zero to find the velocity at t = 2.50 s.

Δxt

## The Attempt at a Solution

(a) was simple, just plug in 2.5s and get 37.5m
(b) the problem doesn't explain well enough, but i finally got it, plug in 2.5 + Δt for t and I got 6(2.5 + Δt)^2
(c) I don't even know how to do this, it's been 4 years since I've taken classes of higher math and I must have forgotten most. The answer provided is 30 but I don't know the steps to get there.
For step c, you need to calculate Δx, which is just the difference in positions of the particle at time t + Δt and at time t. Divide the change in position of the particle, which occurs over the time interval Δt, by Δt and take the limit. Once you do some algebra on this quotient, taking the limit as Δt approaches zero may not be as scary.

I found another problem worked out on this website, so what I did was 6(2.5+Δt)(2.5+Δt) to get 6Δt^2 + 30Δt + 37.5, this equation is x + Δx, so I have to remove the original x of 37.5, leaving me with Δx = 6Δt^2 + 30Δt. take this and put it over Δt to get 6Δt + 30, and plug in 0 for Δt because Δt is going to 0. and I got 30.