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

1. Jan 27, 2012

### bamber296

1. The problem statement, all variables and given/known data
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.

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

3. 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!

2. Jan 27, 2012

### Curious3141

Re: How to determine acceleration from one point to another on a velocity vs time gra

Answer should be 48m.

Remember, distance (which is what's being asked for) is just an unsigned number. If they asked for displacement, then sign (or direction) is important.

Think of it as area under the v-t graph. It's a trapezoidal area, so it's not just base*height. The area under the graph can be calculated as the sum of a triangle (area: 18) and a rectangle (area:30), though.

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.