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NOT SATISFIED WITH THE ANSWER!!!! .. Whatever u have told is already written in book and is self explanatory..... Kindly give me the HOW DO U GET THAT TYPE OF SLOPE FOR THIS TYPE OF ACCELERATION AND THAT TYPE OF ACCELERATION.... KINDLY TELL ME THE APPROACH...!!!! ? ? ? ? ? ? ? ?

Question has been reposted in the answer section of this question... Kindly understand what I am trying to ask!!

see the slope of velocity versus time graph gives you the value of acceleration, so let us now look at the velocity versus time graph.

For 0-10 sec, the slope of the graph is increasing continuously and this implies that the acceleration(slope of v-t) is increasing with time.

At t=0 sec: acceleration=slope=0

Let's calculate acceleration for 0-10sec . For calculating the acceleration, we need slope of velocity versus time graph, so let us pick any two points in the graph and calculate slope, let's take one point as (0,0) and another point as (10,24) then calculate slope which comes out to be 24-0/10-0=2.4 m/s

^{2}

hence at t=10 sec: acceleration = 2.4 m/s

^{2}.

so join these two points by a straight line to get the graph.

Similary in the interval 10-18 S we can see that velocity is constant it is not changing so take any two. Points similarly and calculate slope the slope will come out to be zero.

And for the last part between 18-20s

the slope of the v-t graph is decreasing and constant, so the acceleration will be negetive and constant in this region.

take two points and calculate slope, take the simplest point (20,0) and (18,24) ,slope will be 24-0/18-20=-12m/s

^{2}

Now from 10-18 sec the acceleration was 0.

at 18sec the acceleration is -12m/s2

So the graph directly goes to -12 from 0 at t=18 sec. That's why there is a line connecting 0 and -12.

And since acceleration is constant for 18-20 so acceleration is -12 and then again the acceleration is back to 0m/s

^{2}.

I hope this clears your doubt.

ask again for any query.

regards

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