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Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the time cancels out, and only displacement remains.)
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, s. In calculus terms, the integral of the velocity function v(t) is the displacement function s(t). In the figure, this corresponds to the yellow area under the curve.
Trajectory of a particle with initial position vector r 0 and velocity v 0, subject to constant acceleration a, all three quantities in any direction, and the position r(t) and velocity v(t) after time t. The initial position, initial velocity, and acceleration vectors need not be collinear, and the equations of motion take an almost identical ...
Timing diagram over one revolution for angle, angular velocity, angular acceleration, and angular jerk. Consider a rigid body rotating about a fixed axis in an inertial reference frame. If its angular position as a function of time is θ(t), the angular velocity, acceleration, and jerk can be expressed as follows:
An object with a large projected area relative to its mass, such as a parachute, has a lower terminal velocity than one with a small projected area relative to its mass, such as a dart. In general, for the same shape and material, the terminal velocity of an object increases with size.
Unlike a regular distance-time graph, the distance is displayed on the horizontal axis and time on the vertical axis. Additionally, the time and space units of measurement are chosen in such a way that an object moving at the speed of light is depicted as following a 45° angle to the diagram's axes.
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To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height = / with respect to , that is = / which is zero when = / =. So the maximum height H m a x = v 2 2 g {\displaystyle H_{\mathrm {max} }={v^{2} \over 2g}} is obtained when the projectile is fired straight up.