00:01
Hello everyone we need to compute the distance traveled by the particle in the first six seconds.
00:06
We have given v of t is equal to minus 0 .5 t squared plus 8p.
00:15
So clearly let us integrate this function v of t with respect to the limit 0 to 6.
00:25
So this will be nothing but integral 0 to 6 minus 0 .5 t squared plus 8p into dp so evaluating this we'll get minus 0 .5 t cubed by 3 plus 8 t square by 2 with respect to the limits 0 to 6 so this is nothing but if we substitute the limit we'll get minus 0 .5 into 6 cube by 3 plus 8 into 6 square by 2 minus 0.
01:16
Simplifying this further, we will arrive at minus 36 plus 144.
01:27
So therefore the particles travels 108 8 meters in the first 6 seconds.
01:38
The second part of the question, we are asked to find the distance travelled by the particle during the second six seconds.
01:50
So that will be nothing but d1 is equal to, let us take the second distance as d1, integral 6 to 12, v of t, t, t...