Question

How to read acceleration from a graph and find velocity from area

Original question: 11. The graph at right models acceleration a. What is the acceleration at t=2sect=2sec? b. When is the acceleration 10ft/sec210\,ft/sec^2? c. What is the minimum acceleration? d. If the initial velocity is zero, what is the velocity at t=6sect=6\,sec? e. If the initial velocity is 20ft/sec20\,ft/sec, what is the velocity at t=6sect=6\,sec? f. If the initial velocity is 20ft/sec20\,ft/sec, what is the velocity at t=7sect=7\,sec?

acceleration in (ft/sec2)(ft/sec^2) a(t)a(t) 1515 1010 55 00 5-5 11 22 33 44 55 66 77 tt time (in sec)

Expert Verified Solution

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Key concept: On these problems, the graph is doing most of the work. The key is to separate two ideas: the value of acceleration at a specific time, and the accumulated change in velocity over an interval.

Step by step

For an acceleration graph a(t)a(t):

  • The value of acceleration at a time is just the y-value of the graph.
  • Velocity comes from area under the acceleration curve:

v(t)=v(0)+0ta(s)ds.v(t)=v(0)+\int_0^t a(s)\,ds.

a) Acceleration at t=2t=2 sec

Read the graph value directly at t=2t=2.

b) When is the acceleration 10ft/sec210\,\text{ft/sec}^2?

Find the time values where the graph has height 10.

c) Minimum acceleration

Look for the lowest y-value on the graph.

d) If initial velocity is zero, what is the velocity at t=6t=6 sec?

Use

v(6)=0+06a(t)dt.v(6)=0+\int_0^6 a(t)\,dt.

So the answer is the signed area from 0 to 6.

e) If initial velocity is 20ft/sec20\,\text{ft/sec}, what is the velocity at t=6t=6 sec?

v(6)=20+06a(t)dt.v(6)=20+\int_0^6 a(t)\,dt.

f) If initial velocity is 20ft/sec20\,\text{ft/sec}, what is the velocity at t=7t=7 sec?

v(7)=20+07a(t)dt.v(7)=20+\int_0^7 a(t)\,dt.

What to watch for

Positive area increases velocity; negative area decreases it. If the graph has straight-line pieces, break the interval into simple shapes and add their signed areas.

Pitfall alert

A frequent error is treating acceleration values as if they were velocities. They are not the same thing. Another mistake is adding only the positive regions under the graph and forgetting that regions below the axis subtract from velocity. For a graph-based question, signed area matters more than raw area.

Try different conditions

If the question asked for displacement instead of velocity, you would integrate velocity, not acceleration. If the initial velocity were not given, you would need that starting value before the final velocity could be found. And if the graph were a piecewise constant function, the area calculation would usually become a set of rectangles rather than triangles or trapezoids.

Further reading

signed area, initial velocity, acceleration graph

FAQ

How do you find velocity from an acceleration graph?

Use v(t)=v(0)+∫_0^t a(s)ds. The change in velocity is the signed area under the acceleration graph.

What does a negative area under acceleration mean?

It means the velocity decreases over that interval, because acceleration is acting opposite the positive direction.

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