Answer:
To have a constant velocity, an object must have a constant speed in a constant direction. Constant direction constrains the object to motion in a straight path thus, a constant velocity means motion in a straight line at a constant speed.
Explanation:
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.
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.
{\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
{\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).
{\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.
Relationship to acceleration
Although velocity is defined as the rate of change of position,
{\displaystyle {\boldsymbol {a}}={\frac {d{\boldsymbol {v}}}{d{\mathit {t}}}}.}{\boldsymbol {a}}={\frac {d{\boldsymbol {v}}}{d{\mathit {t}}}}.
From there, we can obtain an expression for velocity as the area under an a(t) acceleration vs. time graph. As above, this is done using the concept of the integral:
{\displaystyle {\boldsymbol {v}}=\int {\boldsymbol {a}}\ d{\mathit {t}}.}{\displaystyle {\boldsymbol {v}}=\int {\boldsymbol {a}}\ d{\mathit {t}}.}
Constant acceleration
{\displaystyle {\boldsymbol {v}}={\boldsymbol {u}}+{\boldsymbol {a}}t}{\boldsymbol {v}}={\boldsymbol {u}}+{\boldsymbol {a}}t
with v as the velocity at time t and u as the velocity at time t = 0. By combining this equation with the suvat equation x = ut + at2/2, i
{\displaystyle {\boldsymbol {x}}={\frac {({\boldsymbol {u}}+{\boldsymbol {v}})}{2}}{\mathit {t}}={\boldsymbol {\bar {v}}}{\mathit {t}}}{\boldsymbol {x}}={\frac {({\boldsymbol {u}}+{\boldsymbol {v}})}{2}}{\mathit {t}}={\boldsymbol {\bar {v}}}{\mathit {t}}.
{\displaystyle v^{2}={\boldsymbol {v}}\cdot {\boldsymbol {v}}=({\boldsymbol {u}}+{\boldsymbol {a}}t)\cdot ({\boldsymbol {u}}+{\boldsymbol {a}}t)=u^{2}+2t({\boldsymbol {a}}\cdot {\boldsymbol {u}})+a^{2}t^{2}}v^{2}={\boldsymbol {v}}\cdot {\boldsymbol {v}}=({\boldsymbol {u}}+{\boldsymbol {a}}t)\cdot ({\boldsymbol {u}}+{\boldsymbol {a}}t)=u^{2}+2t({\boldsymbol {a}}\cdot {\boldsymbol {u}})+a^{2}t^{2}
{\displaystyle (2{\boldsymbol {a}})\cdot {\boldsymbol {x}}=(2{\boldsymbol {a}})\cdot ({\boldsymbol {u}}t+{\frac {1}{2}}{\boldsymbol {a}}t^{2})=2t({\boldsymbol {a}}\cdot {\boldsymbol {u}})+a^{2}t^{2}=v^{2}-u^{2}}(2{\boldsymbol {a}})\cdot {\boldsymbol {x}}=(2{\boldsymbol {a}})\cdot ({\boldsymbol {u}}t+{\frac {1}{2}}{\boldsymbol {a}}t^{2})=2t({\boldsymbol {a}}\cdot {\boldsymbol {u}})+a^{2}t^{2}=v^{2}-u^{2}
{\displaystyle \therefore v^{2}=u^{2}+2({\boldsymbol {a}}\cdot {\boldsymbol {x}})}\therefore v^{2}=u^{2}+2({\boldsymbol {a}}\cdot {\boldsymbol {x}})
