Applying Calculus to Kinematics Problems

A running app records one thing the whole way round: where you are, with a time against it. Positions and a clock, over and over, for the entire route. Nobody strapped a speedometer to your trainers. Even so, the app hands you a speed at every point of that run, a pace per kilometre, and the hill where you slowed down. It can do that because speed is nothing more than how fast your position is changing. Differentiate a distance equation and you get speed. Differentiate the speed and you get acceleration. One equation for where a thing is carries the whole story of how it moves.

This is video six of six in Functions and Differentiation. If today feels shaky, M A fourteen, oh five, Stationary Points: Maxima and Minima, covers the step just before this one.

Displacement, velocity, acceleration. Stack those three as a ladder, because every question in this topic is standing on one of its rungs. The top rung is displacement, written as s. It is where the object is, measured from wherever it started, changing as time goes on. That equation is almost always what a question hands you. Differentiate it with respect to time and you land on the middle rung, velocity. That is not a rule someone invented. Velocity is the rate at which position changes, and differentiating with respect to time is how you find the rate at which anything changes. So d s by d t is not a route to velocity. It is what velocity means, written down in symbols. One word is worth pinning down. Velocity is not quite the same as speed, because velocity carries a direction, and that direction is what the plus or minus sign on your answer is telling you. Do it again and you drop to the bottom rung. Acceleration is the rate at which velocity changes, so differentiating your velocity expression with respect to time gives acceleration, written d v by d t. Carry this line into the exam hall, and say it to yourself as you read the question. One down: velocity. Two down: acceleration. One boundary before the method. If a problem hands you an acceleration that never changes, and no equation to differentiate, this is not the technique it wants: that is a separate method with its own video. Calculus is here because the rate itself is moving. And if you are handed a distance-time graph with no equation, you would measure the gradient off the page by hand. That is a different skill, owned by Estimating the Gradient of a Curve, in the chapter on real-world graphs and rates of change. Quick check, and it is the whole idea in one question. A drone's height above the ground is modelled by an equation in t, and the question asks for its acceleration. Do you differentiate that height equation once, twice, or not at all? Take your time. I'll wait right here. The answer is twice. Height is a displacement, so one differentiation gives the drone's velocity, and a second gives its acceleration. Stop after one and you have answered a different question from the one on the paper. That ladder is the map, and everything from here is walking down it with real numbers on it.

Time for a real question, with a car, a stopwatch and four seconds of motion. A remote control car's distance from its start point, in metres, t seconds after starting, is modelled by s equals t cubed, minus six t squared, plus nine t, for t from zero to four. Find an expression for its velocity, and find the times at which the car is momentarily stationary. The differentiating itself is the power rule applied term by term, which the video on differentiating powers of x builds from the ground up. Here it is in brief. Velocity is one rung down, so differentiate s with respect to t. The t cubed gives three t squared. The minus six t squared gives minus twelve t. The nine t gives nine. So v equals d s by d t equals three t squared, minus twelve t, plus nine. Now the second half. Momentarily stationary means the car is not moving, just for an instant, and not moving means its velocity is zero. So those physical words point straight at a piece of algebra: set the velocity expression equal to zero and solve. That is the same move as finding a stationary point on a curve, which is why the wording sounds familiar. The difference is that this zero has something you can watch: the car has stopped dead. Your turn on the solving. Set three t squared, minus twelve t, plus nine equal to zero, and find both values of t. Have a go at this one. I'll wait. The answer is t equals one and t equals three. Divide the whole equation by three, leaving t squared, minus four t, plus three equals zero. That factorises into t minus one, times t minus three. Both roots sit inside the zero to four seconds the model covers, so both are real moments in this car's journey. Write it as t equals one second and t equals three seconds, units on, and give both. The question asked for times, in the plural, and one of the two alone is half an answer. Now the part a purely mechanical answer walks past. Ask what the car is doing between those two stops. Put t equals two into three t squared, minus twelve t, plus nine, and you get twelve, minus twenty four, plus nine, which is minus three. Minus three is not an error to tidy away. That minus is a direction. The car is travelling back towards the point it set off from, at three metres per second. So the whole four seconds reads off those signs. The car drives away, stops at one second, reverses back to its starting point, stops again at three seconds, then drives away again. And if the question had asked for acceleration, you take one more rung down. Differentiate the velocity: three t squared gives six t, minus twelve t gives minus twelve, and the plus nine gives zero. So acceleration equals six t, minus twelve. Read that sign too. For the first two seconds, six t minus twelve is negative, so the velocity is falling the whole time, from plus nine down to minus three. After two seconds it turns positive and the velocity climbs back up. One equation for position gave you the car's velocity, its direction, its stops and its acceleration.

Nothing in that method was actually about motion, so watch the same two moves work on water. The volume of water in a leaking tank, in litres, t minutes after the leak starts, is modelled by capital V equals five hundred, minus four t squared. Find the rate at which the volume is decreasing after five minutes. There is no distance here and no velocity. There is a quantity changing as time passes, and a question about how fast it is changing, which is the same question in different clothing. So take the whole thing this time. Differentiate capital V with respect to t, then put t equals five into what you get. Pause here and work it through. I'll wait. Here is the working. Five hundred is a plain number, so it differentiates to zero. Minus four t squared gives minus eight t. So d capital V by d t equals minus eight t, and putting five in gives minus forty. Minus forty is the rate at which the volume is changing. It is negative because the volume is going down, so that minus sign is the leak itself. Then the step that decides whether the answer is finished. The question asked for the rate at which the volume is decreasing, and the word decreasing already carries the minus. So the answer is forty litres per minute, units on, with no minus sign in front. Writing minus forty litres per minute decreasing says the same thing twice, and ends up meaning the opposite. Match your final line to the word the question used, and put the units on it. Volume, temperature, cost, population: if an equation gives a quantity against time, differentiating gives the rate that quantity is changing.

Switch into exam mode, because there are three places where working that is completely right still loses its answer. The first is stopping a rung too early. The question asks for acceleration, the working differentiates once, and the velocity goes on the answer line. Read the word in the question before the pen moves, and count the rungs you need. The second is the interval. These models usually arrive with a range of t attached, like zero to four seconds, and a solution outside it is not a time the model describes. Solve the equation, then hold every root against the range you were given. The third is the finish. A rate of change is a quantity per unit of time, so it needs its units: metres per second, litres per minute, degrees per hour. And when the number comes out negative, say what that negative means here. Take a question like this: a tank is losing water, and you are asked how fast the volume is decreasing. Forty litres per minute is an answer. A bare minus forty is a number, and it leaves the interpreting to the person marking it. Learn the wordings that mean differentiate in these questions. Find the velocity. Find the rate at which it is changing. Find when the object is momentarily at rest, or instantaneously at rest. Each of those asks about a rate rather than an amount. Right rung, right range, right units, and the working you already did survives to the answer line.

Time to gather this up, and the ladder does most of the gathering for you. Displacement sits at the top, velocity in the middle, acceleration at the bottom, with one differentiation on each arrow between them. One down gives velocity, and two down give acceleration. Setting the velocity expression equal to zero finds the moments an object is momentarily at rest, because a velocity of zero is exactly what not moving means. Signs are information rather than mess. A negative velocity is movement back towards the start, and a negative rate is a quantity falling. And those two moves work anywhere a quantity is modelled against time, so finish with the units on, and with the word the question actually asked for.

That completes our chapter on Functions and Differentiation. If this one landed, give it a thumbs up so you can see it is done; if not, save it and come back to it. The next chapter is Coordinates and Straight-Line Graphs.

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