00:01
Hi, in the first support of this question, we'll write the even equation for y as 1, sine, 5 by 12x plus 3, 5 .t.
00:18
We know that p is equal to d ,y by dt.
00:25
So that is equal to d by dt of 1, sine 5 by 12x plus 3 pi t.
00:36
From this we can write 1 into 3 pi into cos 5 by 12 x plus 3 pi t.
00:56
Now from the question we need to find out the transfer speed at t is equal to 0 .110 second and x is equal to 1 .40 second and x is equal to 1 .40 meter.
01:17
So v is equal to 1 into 3 pi into cos 5 by 12 into 1 .40 plus 3 pi into 0 .10.
01:40
And upon solving this we will get v is equal to 9 .42 meter per second as the value for the transverse speed.
01:54
So that is the answer for the first separat.
01:57
Part of this question.
02:01
Now in the second subpart, acceleration is equal to dv by d t, that is equal to d by d t of from the first subpart we have v as 1 into 3 pi into cost 5 by 12 x plus 3 pi t and the value of x and t remains the same.
02:30
So, a is equal to 1 into 3 pi into upon derivating this we will get minus 3 pi into sine 5 by 12 into x is 1 .40 meters plus 3 pi into t is 0 .110 seconds.
03:01
And upon solving this, we will get a is equal to minus 2 .18 meter per second square as the transverse acceleration...