00:01
Hi, today we are solving the question in which we are given that the position of a moving hockey puck after t seconds is s t is equals to 10 inverse s where s is in meters.
00:15
So here by the inverse of trigonometric function from here, s -dh -t can be 1 by 1 plus t squared because we know that 10 inverse d by d x of 10 inverse is 1 by 1 plus x square.
01:02
So from here s dash t is equals to 1 by 1 plus t square as the given function is in terms of s in t.
01:12
So here we have to find the velocity of the hockey puck at any time t.
01:17
So firstly hockey puck velocity will be as s is given as position.
01:26
So velocity will be equal to the position here that is equals to 1 .1.
01:32
By 1 plus t square.
01:35
So here velocity is 1 by 1 plus t square.
01:40
Similarly, find the acceleration.
01:45
So we have to find the acceleration.
01:48
So here, at is equals to minus 2t by 1 plus t square, pole square as 80 is equal to minus 2t by 1 plus t square, pole square as a t is equal to differential of vt.
02:09
So we get a t is equals to minus 2t by 1 plus t square whole square.
02:16
Now we are given it.
02:19
We have to evaluate acceleration and velocity for t is equals to 1 to 6 seconds...