Velocity definition and acceleration definition in graph
first the basic point is describing about graph so that you can get idea about graph, i will provide you trick in the last section don’t worry in the above picture example how to construct graph from given problem. you must remember in (x-t) graph this is displacement verses time graph and slope of this graph is define as
Slope = (y2-y1)/(x2-x1) = (x2-x1)/(t2-t1 ) = ∆x/∆t = v = tan𝛳
here on y axis displacement and in x axis time.
as slope will be greater velocity will be grater because slope is equal to velocity in (x-t) in above figure shown velocity is greater slope is also greater important point slope of (x-t) gives velocity if slope is less than 90 degree slope is positive and hence velocity is positive angle always taken from positive x direction we know that to find slope of any curve at any point we draw a tangent to curve on that point and slope of that tangent line give slope of curve at that point hence two basic important point for (x-t) graph if slope is increasing then velocity is increasing and if slope is decreasing means velocity is decreasing and second point is angle if angle say 𝛳< 90 degree then velocity is positive and if 𝛳>90 degree then velocity is negative angle always taken from positive x axis this is already shown in above picture velocity increasing and decreasing case this is actual velocity definition how velocity how velocity increases and decreases and direction changes if 𝛳 is constant its mean that velocity is constant after knowing this basic point now we will do some question see picture below Velocity definition and acceleration definition
here two question for (x-t) graph lets solve one by one take first problem analysis as from question we have to find velocity in region OA, AB, BC.
we know that in (x-t) graph slope gives velocity hence slope of OA will give velocity in region OA we also know that slope is equal to
Slope of OA = (y2-y1)/(x2-x1) = (10-0)/(5-0) = 10/5 = 2 m/s and angle of OA is less than 90 degree hence positive so velocity in region OA is 2 m/s constant in positive direction now similarly for region AB slope put the value of y coordinate and x coordinate slope = (10-10)/(15-5) = 0/10 = 0 m/s hence velocity in AB region is 0 m/s its means that body is at rest no displacement also seen in picture body has no any displacement during region AB now for similarly BC region put the value of y coordinate and x coordinate in slope equation hence slope = (15-10)/(20-15) = 5/5 = 1 m/s slope for BC is less than 90 degree hence positive now you can analyse as velocity in OA region is greater than velocity in region BC hence slope of OA is greater than slope of BC it is clear from figure.
Now for second question see figure trapezoid similarly we can calculate velocity so velocity of region AB = slope = (10-0)/(5-0) = 10/5 = 2 m/s
velocity of region AB = slope = 0 now velocity of region BC
slope = (0-10)/(20-15) = -10/5 = -2 m/s here it is clear that slope of BC is greater than 90 degree so velocity is negative so it is very clear that in (x-t) graph find slope for velocity magnitude and find angle for velocity sign hence for any type of graph in (x-t) you can easily find out velocity in case of curve find velocity with same process but see whether slope is increasing or decreasing if slope is increasing then then velocity is increasing and if slope is decreasing then velocity is decreasing for sign angle less than 90 degree positive and if greater than 90 degree negative this is basic fundamental point for (x-t) graph with the help of this concept you can solve any problem in (x-t) graph.now it is clear that in (x-t) graph either it will be straight line or curve or both so for all cases the process for checking velocity and its sign will be same now for curve velocity will be variable hence acceleration will be there so for acceleration sign just check velocity is increasing or decreasing if velocity is increasing then acceleration will be positive if velocity is decreasing then acceleration will be negative second method to check acceleration sign if concave of graph is up then acceleration will be positive if concave of graph is down then acceleration will be negative it is important point for acceleration sign.
In (v-t) graph slope of curve gives acceleration as in above figure given now for acceleration positive or negative same process for angle means slope is used and area under this graph gives displacement this is two important point for (v-t) graph now from Q1 we have to find acceleration in region OA, AB and also displacement lets solve first we will find acceleration hence slope of OA = (10-0)/(5-0) = 10/5 = 2 m/s² and angle is less than 90 degree so positive now for acceleration in region AB it is clearly seen that velocity is constant means slope is zero but how slope of AB = (10-10)/(10-5) = 0/5 = 0 so acceleration in region AB is 0 now for displacement during OA we know that for displacement find area under curve (v-t) here it is triangle so area of tingle = 1/2*5*10 = 25 m now for displacement during AB it is rectangle hence area of rectangle = 10*5 = 50 m so for total displacement during OAB = 25+50 = 75 m
now for Q2 find displacement and distance we know that area under curve (v-t) gives displacement but remember above x axis area will be positive and below x axis it will be negative now you can find area and displacement hence above x axis area = displacement = 1/2*5*10 = 25 m similarly below x axis area = 1/2*5*10 = 25 m but it is negative hence total displacement = 25-25 = 0 now for distance both will be added = 25+25 = 50 m because weather body moved forward or backward distance will be total addition.
now from this concept you can convert (x-t) graph to (v-t) and vice versa using concept of both graph now last graph is (a-t) graph
(a-t) graph slope gives Jerk that is da/dt = Jerk m/s³ area of this graph gives change in velocity.
Trick for solving graph questions.
(x-t) graph ↓
(v-t) graph (Slope)
remember this series from top to bottom means to say that if (x-t) graph is given and asking for velocity so find slope if (v-t) graph given and asking for acceleration find slope similarly
(v-t) graph (Area)
(a-t) graph ↑
remember this series from bottom to top means to say if (a-t) graph is given and asking for velocity then find area if (v-t) graph is given and asking for displacement and distance find area.
Now i hope that you have enjoyed to learn Velocity definition and acceleration definition and you have learn the concept for graph thanks for reading we will continue in next post.
dated 7th July 2018.