Question

formula for velocity

Answer 1

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HERE IS YOUR ANSWER

✍️ Equation for velocity.

✍️ The equation and formula for velocity is similar to speed.

✍️ To figure out the velocity you divide the distance with time it takes to travel that same distance than you add you direction to it.

$velocity = \frac{displacement}{time}$

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Answer 2

Answer:

Explanation:

Average velocity

Velocity is defined as the rate of change of position with respect to time, which may also be referred to as the instantaneous velocity to emphasize the distinction from the average velocity. In some applications the "average velocity" of an object might be needed, that is to say, the constant velocity that would provide the same resultant displacement as a variable velocity in the same time interval, v(t), over some time period Δt. Average velocity can be calculated as:

${\displaystyle {\boldsymbol {\bar {v}}}={\frac {\Delta {\boldsymbol {x}}}{\Delta {\mathit {t}}}}.}{\boldsymbol {\bar {v}}}={\frac {\Delta {\boldsymbol {x}}}{\Delta {\mathit {t}}}}.$

The average velocity is always less than or equal to the average speed of an object. This can be seen by realizing that while distance is always strictly increasing, displacement can increase or decrease in magnitude as well as change direction.

In terms of a displacement-time (x vs. t) graph, the instantaneous velocity (or, simply, velocity) can be thought of as the slope of the tangent line to the curve at any point, and the average velocity as the slope of the secant line between two points with t coordinates equal to the boundaries of the time period for the average velocity.

The average velocity is the same as the velocity averaged over time – that is to say, its time-weighted average, which may be calculated as the time integral of the velocity:

${\displaystyle {\boldsymbol {\bar {v}}}={1 \over t_{1}-t_{0}}\int _{t_{0}}^{t_{1}}{\boldsymbol {v}}(t)\ dt,}{\boldsymbol {\bar {v}}}={1 \over t_{1}-t_{0}}\int _{t_{0}}^{t_{1}}{\boldsymbol {v}}(t)\ dt,$

where we may identify

${\displaystyle \Delta {\boldsymbol {x}}=\int _{t_{0}}^{t_{1}}{\boldsymbol {v}}(t)\ dt}\Delta {\boldsymbol {x}}=\int _{t_{0}}^{t_{1}}{\boldsymbol {v}}(t)\ dt$

and

${\displaystyle \Delta t=t_{1}-t_{0}.}\Delta t=t_{1}-t_{0}.$

Instantaneous velocity

Example of a velocity vs. time graph, and the relationship between velocity v on the y-axis, acceleration a (the three green tangent lines represent the values for acceleration at different points along the curve) and displacement s (the yellow area under the curve.)

If we consider v as velocity and x as the displacement (change in position) vector, then we can express the (instantaneous) velocity of a particle or object, at any particular time t, as the derivative of the position with respect to time:

${\displaystyle {\boldsymbol {v}}=\lim _{{\Delta t}\to 0}{\frac {\Delta {\boldsymbol {x}}}{\Delta t}}={\frac {d{\boldsymbol {x}}}{d{\mathit {t}}}}.}{\boldsymbol {v}}=\lim _{{\Delta t}\to 0}{\frac {\Delta {\boldsymbol {x}}}{\Delta t}}={\frac {d{\boldsymbol {x}}}{d{\mathit {t}}}}.$

From this derivative equation, in the one-dimensional case it can be seen that the area under a velocity vs. time (v vs. t graph) is the displacement, x. In calculus terms, the integral of the velocity function v(t) is the displacement function x(t). In the figure, this corresponds to the yellow area under the curve labeled s (s being an alternative notation for displacement).

${\displaystyle {\boldsymbol {x}}=\int {\boldsymbol {v}}\ d{\mathit {t}}.}{\displaystyle {\boldsymbol {x}}=\int {\boldsymbol {v}}\ d{\mathit {t}}.}$

Since the derivative of the position with respect to time gives the change in position (in metres) divided by the change in time (in seconds), velocity is measured in metres per second (m/s). Although the concept of an instantaneous velocity might at first seem counter-intuitive, it may be thought of as the velocity that the object would continue to travel at if it stopped accelerating at that moment.

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