00:01
For this problem, we're being asked a series of questions regarding a rock being thrown up the surface of the moon.
00:08
And this equation here gives the position in respect to time, position in meters in respect to time in seconds.
00:17
So for part a, it's asking us, basically, what is this equation in the form for, in terms of velocity and acceleration, right? so to find velocity, the velocity of this rock being thrown up, given the equation here, what we do is we need to take the derivative of this equation because the derivative position is velocity.
00:36
So to do that, we get at the end 24 minus 1 .6 t.
00:43
This is our velocity function, velocity of the end meters per second.
00:49
Okay.
00:51
And as for acceleration, acceleration is essentially the derivative of velocity.
00:55
So we take the derivative of this value to get acceleration, and we get that acceleration is negative 1 .6 meters per second squared, right? and the one point, the negative indicates towards the moon.
01:13
So yes, okay.
01:15
Moving on to part b, the question asks, at what time will this object be at the peak of its trajectory? so to find this, you have to recognize that when you throw an object up, it reaches the peak over here when velocity equals zero.
01:34
So what we can do is we can take this equation here and set it equals zero and find the time.
01:39
So our equation for ds is 24 minus 1 .6t and we set this equal to zero because velocity equals zero at the peak, right? then we get that t equals 15 seconds.
01:56
They're ugly five, 15 seconds at the peak.
02:03
All right, and moving on to part c, it's asking us, so now that we have found this, what is the max height of our object? so we know that the peak will be at t equals 15 seconds.
02:17
What we can do is we can take this value, we can plug it back into our position function right here, right? so position function is s equals 24, 5 ,000 minus 0 .0 .0.
02:29
8 multiplied by 15 squared.
02:32
And after you simplify this value, you get that the maximum height is 180 meters.
02:41
Ok, moving on to part d.
02:43
So part d asks, so now that you have the maximum height, what is the time, how long does it take for the object to reach half of this volume? so we're given this function here, right? so s equals 24t minus 0 .8 t squared.
03:02
If you want to find the time in which the peak is half, we can essentially set this position function equal to half of the maximum height, right? half of 180 would be 90 in this case, right? and then we would solve for this.
03:17
So simplifying this over into a friendly quadratic, we get that this essentially becomes 0 .8t squared, minus 204t plus 19...