Solving a Cart Distance Problem: 18s to 21s

[bamber296]

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=(v²_f−v²_i)/2a
Δx=0.5at²
Δx=0.5(base)(height)

The Attempt at a Solution

I used the first equation as:
Δx=(v²_f−v²_i)\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 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=(v²_f−v²_i)/2a
Δx=0.5at²
Δx=0.5(base)(height)

The Attempt at a Solution

I used the first equation as:
Δx=(v²_f−v²_i)\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!

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.