00:01
In this part, for part a, in the given time interval from 0 to 4 .0 seconds, the acceleration of object, which is equals to 0 meter per second squared, the reason is that since in the above time interval, the velocity of object is constant.
00:27
For part b in time interval your 4 .0 seconds to 9 .0 seconds the acceleration change is change in velocity divided by time interval is equal to dv over d t is equals to v.
00:52
V .1.
00:52
V .1 divided by time interval is equal to dv over d t is equals to v squared minus v1 divided by t2 minus t1 so at t1 we have 4 .0 seconds your v1 is negative 12 .0 meter per seconds your t2 is 9 .0 seconds and your v2 is 18 .0 meter per seconds so a is equal to 18 .0 minus negative 12 .0 divided by 9 .0 0 minus 4 .0 so your acceleration will be equal to 6 .6 .6 meter per second squared.
01:44
For part c we need to know the acceleration so in time interval we have 13 .0 seconds to 18 .0 seconds so still the same acceleration is equal to change in velocity divided by time interval so definitely our t1 is 13 .0 seconds.
02:08
Our v1 is 18 .0 meter per seconds.
02:13
Our t2 is equals to 18 .0 seconds.
02:18
Our v2 is equal to 0 meter per seconds.
02:21
So a is equal to 0 minus 18 .0 rather.
02:27
So we have 18 .0 divided by 18 .0.
02:37
Minus 13 .0...