# How to determine acceleration from one point to another on a velocity vs time graph

## Homework Statement

Evaluate the distance the cart travels from 18.0s to 21.0s.

The points are roughly (18.0, 22) and (21.0, 10). The slope/acceleration is constant.

## Homework Equations

Δx=(v2f−v2i)/2a
Δx=0.5at2
Δx=0.5(base)(height)

## The Attempt at a Solution

I used the first equation as:
Δx=(v2f−v2i)$\overline{}2a$
=(222−102)/2(-4) = -48 m

Δx=0.5at2 = 0.5(-4)(9) = -18 m

Both of those were wrong along with the last equation (which also equaled -18 m). I really need help!

Curious3141
Homework Helper

## Homework Statement

Evaluate the distance the cart travels from 18.0s to 21.0s.

The points are roughly (18.0, 22) and (21.0, 10). The slope/acceleration is constant.

## Homework Equations

Δx=(v2f−v2i)/2a
Δx=0.5at2
Δx=0.5(base)(height)

## The Attempt at a Solution

I used the first equation as:
Δx=(v2f−v2i)$\overline{}2a$
=(222−102)/2(-4) = -48 m

Δx=0.5at2 = 0.5(-4)(9) = -18 m

Both of those were wrong along with the last equation (which also equaled -18 m). I really need help!

What you worked out with $v_f^2 = v_i^2 + 2a{\Delta}x$ is essentially correct, except the final velocity here is 10m/s and the initial velocity is 22m/s. So you wouldn't have got a negative sign if you'd arranged it properly.
The $\frac{1}{2}at^2$ only applies if the object is starting from rest.