00:01
Hello students, in this question we have a person starting with initial velocity point position which is 8 meter and the initial velocity which is given as 3 .2 meter per second and the acceleration is equal to 0 .8 meter per second square.
00:24
So, we need to first thing we need to find the slope of the intercept of velocity versus time graph.
00:32
So, we can write down that the equation v is equal to u plus 8t or we can write down the position equation which is s is equal to s0 plus ut plus half 8t square.
00:47
So, let's substitute the value.
00:49
So, s the position final position s will be equal to s0 which is the initial position that is 8 plus u which is the initial velocity which is 3 .2t plus half times 0 .8t square.
01:07
This is the initial position.
01:10
So, we can find out the final velocity ds by dt which is equal to final velocity v is equal to 3 .2 plus half 0 .8t times 2.
01:31
So, 2 and 2 will cancel.
01:32
So, that is just v is equal to u plus 8t.
01:35
So, this is the equation right.
01:37
So, we can see clearly see that the velocity equation v of time.
01:42
So, v of t is equal to 3 .2 plus 0 .8t.
01:47
So, we can plot the graph with v and time...