Physics

From a particle's velocity-time graph, its average velocity can be found by calculating the total area under the graph and then dividing it by the corresponding time-interval.

For a particle with uniform acceleration, velocity-time graph is a straight line. Its average velocity is given by $v_{avg}=(v_{i}+v_{f})/2$. From the graph, this can be found by drawing the y-intercepts of initial and final velocities and then drawing the mid-point.

In the given graph, instantaneous velocity $=v_{1}$ at $t=t_{1}$ and $v_{2}$ at $t=t_{2}$.

Average velocity is found at E, i.e. mid-point of C and D.

$a=t_{2}−t_{1}v_{2}−v_{1} $

In the given graph,

$a=4−240−20 =10m/s_{2}$

Acceleration may be positive, negative or zero as it is a vector quantity.

In the given graph, displacement is found as area of trapezium ABCD.

$s=21 ×DC×(AD+BC)$

Note: Since, displacement is a vector quantity, Area below the time-axis is considered as negative and above it is considered as positive.

as $a=dtdv $

or $dv=adt$

or $Velocity=∫adt$ (Which is area under the curve.)

In case where the acceleration is zero, tthe slope is zero ( horizontal line). If the acceleration is positive, then the slope is positive (an upward sloping line).