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  2. Why is displacement equal to the area of velocity-time graph?

    physics.stackexchange.com/questions/415414/why-is-displacement-equal-to-the...

    the general formula of speed (v) is v=distance (s)÷time taken (t) so the formula of distance (s) should be s=v×t so if the speed-time graph of a body is a triangle, the distance covered by the body should be equal to twice of the area of the graph because area of a triangle is - 1/2×base×height where base=t and height =v so it becomes 1/2× ...

  3. kinematics - Does the area under a velocity-time graph represent...

    physics.stackexchange.com/questions/692111/does-the-area-under-a-velocity-time...

    If the velocity time graph depicts the magnitude of velocity (which can also be a vectorial magnitude), the area under the curve is distance (arc length): $$\mbox{distance}=\int |\vec v(t)| \mbox{d}t$$ If velocity is signed velocity (which can be generalized to a velocity vector), then the area (or areas) is displacement.

  4. kinematics - How area under Velocity-Time graph represents...

    physics.stackexchange.com/questions/344837

    (1). velocity=displacement/time (2). displacement=velocity $\times$ time =area of rectangle =Velocity-axis $\times$ time-axis, putting the value of velocity and time, then getting out the displacement. But what if the acceleration is constant instead of $0$ (where graph shows a triangle) ?

  5. Distance traveled equals area under velocity time graph

    physics.stackexchange.com/questions/376115

    2. The relationship you are referring to is. v = Δx Δt, v = Δ x Δ t, which gives you the area under the velocity curve (the displacement) when you rewrite it as Δx = vΔt Δ x = v Δ t. This is the discrete form of this relationship; it only works for large time intervals. If you wanted to know the velocity at any instant, you have to take ...

  6. The areas under velocity-time graph a particle in a given ... -...

    www.toppr.com/ask/question/the-areas-under-velocitytime-graph-for-a-particle...

    The velocity time graph of a particle along a straight line is given in figure. In the time interval t = 0 to t = 8 seconds, answer the following questions: The value of [ m a g n i t u d e o f maximum v e l o c i t y m a g n i t u d e o f a v e r a g e v e l o c i t y for the given interval is

  7. Area under a displacement graph - Physics Stack Exchange

    physics.stackexchange.com/questions/62845

    I think it just represents what you said: the area under a displacement-time graph. I can't think of any other use for it. There are two main reasons for this: Your quantity, let's call it f(t) f (t), retains a memory of where the object has been. That's because the area under the graph depends on the entire past history of the graph, and not ...

  8. The area under velocity-time graph gives: - Toppr

    www.toppr.com/ask/question/the-area-under-velocitytime-graph-gives

    The area enclosed under the velocity-time graph gives acceleration. View Solution. Q2.

  9. Area under velocity-time graph is equal to the: - Toppr

    www.toppr.com/ask/question/area-under-velocitytime-graph-is-equal-to-the-2

    The area under velocity - time graph represents the-. View Solution. Click here:point_up_2:to get an answer to your question :writing_hand:area under velocitytime graph is equal to the 2.

  10. Area under a velocity-time graph gives the: - Toppr

    www.toppr.com/ask/question/area-under-a-velocitytime-graph-gives-the

    (a) If a body moves with uniform velocity, its acceleration is _____. (b) The slope of a distance-time graph indicates _____ of a moving object. (c) The slope of a speed-time graph of a moving body gives its _____. (d) In a speed-time graph, the area enclosed by the speed-time curve and the time axis gives the _____ by the body.

  11. Draw a velocity-time graph an object in uniform motion show ... -...

    www.toppr.com/ask/question/draw-a-velocitytime-graph-for-an-object-in-uniform...

    Draw a velocity-time graph for an object in uniform motion show that the area under the velocity-time graph gives the displacement of the object in the given time interval. View Solution Q 2