Problem 6 - velocity versus time graph

Given is graph of motion of an object moving along a straight line. Determine from this graph the:

a) total distance traveled,

b) average velocity for this motion,

c) maximum and minimum accelerations.

Solution.

Let us use the following notation.

Consecutive values of velocities, which can be easily identified from the graph:

v₀ =0 m/s; v₁=1.0 m/s; v₂ = 3.0 m/s; v₃ = 0.0 m/s

Durations of travel:

t₀ = 0.0 s; t₁ = 1.0 s; t₂ = 2.0 s; t₃ = 4.0 s; t₄ = 7.0 s; t₅ = 9.0 s

Answer to a)

The total distance traveled is equal to area A under the graph. This area can be calculated after dividing it into triangles Tn and rectangles Rm and summing up all partial areas. These areas can be written as:

T1 = (1/2) t₁ v₁

T2 = (1/2) (t₃ – t₂) (v₂ – v₁)

T3 = (1/2) (t₅ – t₄) v₂

R1 = (t₃ – t₁) v₁

R2 = (t₄ – t₃) v₂

Substituting numerical values for times and velocities, which can be read from the graph, and adding these number we get for total area under the graph, that is for the total distance traveled

A = 17.5 m,

(dimension of product tv is meters).

Answer to b)

The average velocity is the ratio of the total distance traveled to the total time of travel

v_av = A / t₅ = 17.5 / 9 = 1.9444 m/s

Answer to c)

There are 3 time spans on the graph, where acceleration is different from zero

Δt₀₁= t₁ – t₀; Δt₂₃ = t₃ – t₂; Δt₄₅ = t₅ – t₄.

The change of velocity in each of these time spans is

Δv₀₁ = v₁ – v₀; Δv₁₂ = v₂ – v₁; Δv₂₃ = v₃ – v₂

By definition, acceleration in linear motion with constant acceleration (within each of time span acceleration is constant) is

a = Δv / Δt

Accelerations in each of the three listed above time regions are

a₁ = (Δv₀₁/ Δt₀₁); a₂ = (Δv₁₂/ Δt₂₃); a₃ = (Δv₂₃/ Δt₄₅).

Substituting the numerical values determined from the graph gives

a₁ =1.0 m/s²; a₂ =1.0 m/s²; a₃ = -1.5 m/s².

The maximum accelerations are a₁ and a₂. The minimum is a₃, which is negative.

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