CALC. A car's velocity as a function of time is given by v_x(t) = α + βt^2, where α = 3.00 m/s and β = 0.100 m/s^3. (c) Draw v_x-t and a_x-t graphs for the car's motion between t = 0 and t = 5.00 s.

Video transcript

Hey, everyone in this problem, we're told that functions are used to represent various changing quantities as functions of position and time. The dependence of the velocity of a rocket on time is given by a function V X of T is equal to A plus B multiplied by T squared If a is equal to 20 m/s And B is equal to 0.25 m/s cubed. We to graph V X T and A X T for the rockets motion between T equals zero and T equals eight seconds. All right. So we're given four answer choices and each of them show a graph of the velocity V X T versus time. And then a second graph for the acceleration A X T as a function of time. Now option a has a velocity starting at zero And increasing somewhat could quadratic to T equals eight seconds to a value of about 16 m/s. And it has the acceleration constant at 0.5 m/s squared throughout the entire eight seconds. Option B A shows the same velocity curve as option A but the acceleration curve is now a straight line. A, a linear line starting at zero m per second at T equals zero seconds all the way up to four m per second squared at eight seconds. Option C Has the velocity as a similar quadratic looking curve. But this time it's starting at 20 m/s and increasing upwards. OK? To about 36 m per second at eight seconds. The acceleration matches the acceleration from option B OK. It's linear. It's a straight line going from zero m per second squared up to four m per second squared. Option D has the same velocity as option C OK. That quadratic looking function starting at 20 m per second and increasing And it has a constant acceleration of 0.5 m/s square throughout the entire eight seconds. Now, we're given the velocity as a function of time. We're given that V X of T is equal to A plus BT squared and we're also given the values of A and B And so we get that V is equal to 20 m/s Plus 0.25 meters per second cubed T squared. Now, this unit meters per second, cubed might look a little strange, but we're multiplying by T squared. OK? If our units of T R seconds, we're multiplying by seconds squared. And so that's gonna give us a unit of meters per second like we want for velocity. Now, this is our velocity, the easiest way to graph something. When we have a function like this, we're just gonna go ahead and make a table of values. We plot out what it should look like. So we have the time T in seconds, we have the velocity V X T starting at T equals zero. OK? And we're just gonna go 123456 all the way up to eight seconds. Yeah, that's the interval we were asked to look at. So we're gonna use our function at T equals zero. We substitute zero into our function. OK? The second term goes to zero and we're left with just 20 m/s. If T is equal to one, we get 20 m per second plus 0.25 m per second, We get 20.2, m/s. If T is equal to two K T squared is going to be equal to four, multiplied by 0.25, gives us one m per second. And then we add our 20 m per second And we get meters per second and we're gonna do the same thing all the way down. If we substitute three into our function, We get 22.25 m/s. You can work this out by doing the calculation yourself by using your calculator, whatever you'd like to do. Subsequuting in four seconds, we get 24 m/s Per five. We get 26.25 m. /s. For T equals six, You get 29 m/s, Per t equals seven, m/s. And for eight, we get 36 m/s. And what you'll notice is that these velocities are getting further and further apart. The higher we go with T, and that's because of this square dependence on T. And we don't have a linear dependence on T, we have a quadratic dependence on T. OK. So this is the table of values for our velocity, gonna give us some more room. We're gonna draw the graph below and this is just gonna be a sketch. So we have our axis here and we're gonna draw the Y axis. OK. The velocity in increments of in our X axis will do All the way up to eight And increments of two. And again, our X axis is the time Now at T equals zero, our function starts at 20 seconds or sorry, 20 m/s. At T equals one. It increases a tiny bit To 20.2. 5 T equals two increases a tiny bit more to 21. And at T equals eight seconds, we get all the way up To 36 m/s. And you can go ahead and draw all of these lines at six seconds. We at 29, all of these dots. When you see that we increase somewhat could quad, it looks quadratic and that makes sense because we have that T squared. So it is in fact quadratic in this way. So our velocity starts at 20 m/s, increases quadratic up to 36 m/s. If we compare to our answer choices, the velocity we found and we described matches choice C or D option A and B have the velocity starting at zero m per second, which we know is not true. OK. So we can already eliminate option A and option B and now we need to look at the acceleration but we were given the function for the velocity, we weren't given the function for the acceleration. So how can we do the same thing or recall that the acceleration A X of T is related to the velocity through the derivative. The acceleration is just equal to the derivative of the velocity with respect to time. If we take the derivative of this velocity function, the derivative of 20 m/s, that's just a constant. So the derivative is gonna be zero 0.25 multiplied by T squared. We take the derivative of T squared. The exponent of two is gonna come down in front and multiply. So we're gonna have .25 multiplied by two which gives us .5. And then the exponent is gonna decrease by one. And so we're gonna get that the acceleration is 0.5 m/s cubed multiplied by the time T And if you look at this as a function of time, OK. This is a linear function. We have T D exponent one. This is linear, we are given the slope 0.5 m per second cubed. And we can write out a table of values as well if we want just to confirm what we've said. So we have 012345678. OK. A linear function with a slope at T equals zero or acceleration is zero, T equals one 0.5, T equals 2.5, multiplied by two, we get one and we can do this over and over with each of these values substituting them in for T multiplying by half. OK. And you can see that this time with the acceleration, these are increasing by the same amount every time because we have that linear term of T instead of the quadratic term of T. If we wanna graph this, the acceleration and the vertical axis, you have the time in seconds along the horizontal axis, We are starting at 00 and we're increasing. Oops, not that much. We are increasing By 0. each time. And I haven't drawn all the points I've drawn kind of every other point. You get the idea that this is a straight line slope, 0.5 Starting from 00. If we go to our answer traces now, OK. We've already narrowed it down to C and D based off the velocity. Yeah, the velocity was starting at 20 m/s and increasing quadratic the acceleration we found starts at 00 and increases linearly till four m per second squared. And so this corresponds with answer choice C and those look just like the graphs we found they match those table of values that we calculated. And so that is the correct answer. Thanks everyone for watching. I hope this video helped see you in the next one.