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v-t Graph

How is averag...

Question

How is average velocity found by using velocity-time graphs?

Average velocity = \frac{Area under v-t graph in the time-interval}{time interval}

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Average velocity = \frac{v (t_{2}) - v (t_{1})}{t_{2} - t_{1}}

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Average velocity = \frac{v (t_{2}) + v (t_{1})}{2}

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None of the above

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Solution

The correct option is C $Average velocity = \frac{Area under v-t graph in the time-interval}{time interval}$
$Average velocity = \frac{Area under v-t graph in the time-interval}{time interval}$
The above formula is valid for all v-t graphs.
Option B and option C are valid only for constant acceleration graphs.

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Similar questions

v_{a v e r a g e} = (1 / 2) (v (t_{1}) + v (t_{2}))

v_{a v e r a g e} = [r (t_{2}) - r (t_{1})] / (t_{2} - t_{1})

v (t) = v (0) + a t

r (t) = r (0) + v (0) t + (1 / 2) a

t^{2}

a_{a v e r a g e} = [v (t_{2}) - v (t_{1})] / (t_{2} - t_{1})

t_{1}

t_{2}

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₂ to t₁?

(a) x(t₂) = x(t₁) + v(t₁)(t₂–t₁) + ()a(t₂–t₁)²

(b) v(t₂)= v(t₁) + a(t₂–t₁)

(d) a_Average = (v(t₂) – v(t₁)) / (t₂ – t₁)

(e) x(t₂) = x(t₁) + v_Average(t₂ – t₁) + ()a_Average(t₂ – t₁)²

(f) x(t₂) – x(t₁) = area under the v–t curve bounded by the t-axis and the dotted line shown.

20 k m

12 k m

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