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

  • Thread starter bamber296
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  • #1
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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)[itex]\overline{}2a[/itex]
=(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!
 

Answers and Replies

  • #2
Curious3141
Homework Helper
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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)[itex]\overline{}2a[/itex]
=(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 [itex]v_f^2 = v_i^2 + 2a{\Delta}x[/itex] 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 [itex]\frac{1}{2}at^2[/itex] only applies if the object is starting from rest.
 

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