00:01
For this question, we have a car that is accelerating from rest at the intersection, at the rate of 3 .2 meters per second square.
00:13
And at the same time, we have a truck that is at the same point, not accelerating, but moving off with a velocity of 20 meters per second.
00:26
So the truck of course will be accelerate pass and overtake the car at first.
00:38
We want to find when the car will move and start to overtake the truck.
00:48
So beyond that, at that point when it overtakes, at this second time, when it overtakes, it is when both the car and the truck actually traveled the same amount of distance, right? this unknown distance over here and also at the same amount of time t.
01:20
So for our truck, the distance d is just 20 meters plus seconds time t times t.
01:30
Whereas for the car, this distance is equals to ut plus half a t square where u is the initial velocity where initial velocity is 0 so we don't have this term we just have half a t square it's just half times 3 .2 times t squared and since they both have traveled the same distance we can equate these two equations together and actually we can quickly solve for t which is 12 .5 seconds and therefore we can find this distance d by just putting in 12 .5 into here so we get 20 times 12 .5 to give us 250 meters now for the next part you'll find how fast the car is traveling when it just overtakes.
02:55
So given that it's constantly accelerating at this rate, so the final velocity will be just aat since it has no initial velocity.
03:04
So the total time is 2 .5, so just 2 .5 seconds times 3 .2 meters per second square and we would have speed of about 40 meters per second.
03:29
Now to sketch the graph for both vehicles.
03:40
I'm gonna start off with the position time graph.
03:47
We start off at the intersection being at the origin over here.
03:53
Now for the truck, it's quite simple because it is constantly moving at a constant velocity...