00:01
In this problem, the acceleration of cathay cool curve is given ak is 4 .9 meter per second square.
00:12
And the acceleration of stan leaves curve is given 3 .50 meter per second square.
00:25
In the part a, we have to find time at which cathy overtakes stand.
00:32
For when they will meet the distance traveled by the catechus and the stand will be same so we can say this distance travel by catechus will be equal to distance travel by stand so it is given in the question that time for a stand is one second more that time for catecules so so x k will be half of ak, tk square, half of as and ts.
01:23
We will know that ts is equal to tk plus 1.
01:26
So if we will put that value here in this equation, as, tk plus 1 whole square.
01:40
Now, ak is 4 .9, as is 3 .5.
01:49
This will be equal to tk plus 1 divided by tk whole square.
01:57
If you solve this, you will find that time takes equal to 5 .46 seconds.
02:08
It is the time for cathay cools to overtake the stand.
02:14
So this is the solution of part a.
02:18
For part b, we have to find the displacement c travels before the before c casin.
02:29
The displacement will be equal to xk is equal to half of a k tk square...