00:01
Hello, here we have to solve the following problem.
00:02
We have to use the graph and first we have to find acceleration at one second.
00:07
So this acceleration is 10 meters per second over 1 .5 seconds.
00:21
That is 6 .67 meters per square second.
00:30
Now in question 2, in question b we have to find the position of an object at 1.
00:38
A half seconds.
00:40
So this position equals to acceleration times t squared over two.
00:54
That's 7 .50 or 7 .5 meters.
00:58
Now in question c we have to find acceleration at two seconds and here as we can notice acceleration is zero meters per square second because velocity is constant.
01:09
In question z we have to find the distance traveled from one and a half to three seconds.
01:17
And in its position at t2.
01:21
So first let's calculate the position at three seconds or probably let's just calculate the distance.
01:34
Oh yeah let's i think it will be easier for us.
01:37
Let's calculate the distance and this distance is one half times um one point five second plus one point five second times 10 meters per second.
01:54
And basically this expression is the area under vt graph...