00:01
This problem we have to find how much is the distance between the deer and the car when the car comes to the rest.
00:08
So the given values are each of the speed of the car is equal to 20 meter percent.
00:13
The final speed is equal to 0 meter per second to rest.
00:19
The distance between the car and the deer is 35 meter and the reaction time to put the trait is c1 equal to 0 .4.
00:33
Seconds.
00:35
What is the distal travel by the card is the reaction pack.
00:39
Let that be d1, which will be given by v1 times t1.
00:45
Now if we put the numerical values, we'll get d1 is equal to 20 times 0 .5, 10 meters.
00:55
The deacceleration a is equal to minus 12 meter per second now the distance travel after applying the break is d2 which will be given by the formula v2 squared minus v1 squared divided by two times acceleration.
01:27
If you put the numerical values we will get d2 is equal to 0 minus 400 divided by 2 times so the distance between the distance between the car and the tier is given by d which is equal to the length l minus distance travel when the break was applied minus during the reaction time.
02:17
It will be equal to 35 and 10 16 .66...