in this video we're looking at a one-dimensional kinematics problem where two objects are moving
and interacting so i have car number one located at x equals zero car number two located at x
equals 200 meters and when the clock starts both cars begin accelerating toward each other
with differing accelerations a1 is 3 meters per second squared a2 is 5.5 meters per second squared
in magnitude but it points to the left so i put a minus sign on it in part a i want to get the time
at which these cars run into each other so the key to these problems is to get a position versus
time function written down for each of these cars and that comes from our basic kinematics formula
x equals x naught plus v not t plus one half a t squared that's describing the
position of an object as a function of time so i write down my position function
for car number one and that's given by the initial position which is zero plus the initial velocity
times time which is a zero times t plus a one half times the acceleration times the
square of the time and those first two terms are gone and that three half simplifies to 1.5
and now i can describe the position of car number one for all moments in time it's just 1.5 times
t squared we do the same thing for car number two but the initial position of car 2 is 200.
the initial velocity is 0 because it started from rest and then my acceleration has a minus
sign in it so i'm going to pull that out in front i get negative one half times 5.5
times t squared and this gives me a 200 minus half of 5.5 which is 2.75 times t squared so now
we have two position functions that describe the position of each of these cars for all
moments in time now we have to do a little bit of physical reasoning what does it mean for the
cars to crash in terms of the mathematics here it means the positions are the same
so x1 is equal to x2 then i can plug in the position functions for each of these cars
x1 as a function of time is 1.5 t squared x2
is 200 minus 2.75 t squared i add the quadratic term to both sides
and this gives me 4.25 t squared is equal to 200 and finally we divide by 4.25 and square
root the result and this gives me a time of 6.86 seconds okay now that i know what time
it is when the cars crash i'm asked in part b to find the position at which they crash
so where are they relative to the origin which i put at the front of car one and you have two
choices of how to do this and it's actually best to do it both ways as a check on your work
i can find the position of car one evaluated at t equals 6.86 seconds so i just plug into my
position as a function of time which was 1.5 times the time squared for that car and when
i run the numbers on this i get 70.6 meters so the crash happens let's say right about here
about 70.6 meters closer to the starting position of car 1 because it wasn't accelerating as fast
and so technically we're done with the problem but i always go back and check the position of the
other object to make sure it agrees so what's the position of card 2 when we plug in 6.86 seconds
that's 200 minus 2.75 times 6.86 squared and this agrees with my first answer giving me 70.6 meters
now i'm very confident i have the right answer if you enjoyed this video or at least found it
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