M1: Kinematics of Motion in a Straight Line | A Level Mathematics (9709) AS Revision Notes

Introduction

Hey! Ready to get your head around motion? Kinematics might sound intense, but it's basically the science of how things move. In this chapter, we're going to build your ultimate toolkit for tackling any motion problem. First, we'll become visual masters with Displacement-Time and Velocity-Time graphs, learning how to read the story of a journey just by looking at its shape. Then, we'll unlock the five 'magic' SUVAT equations – your go-to for any situation where acceleration is constant. Finally, for those trickier problems where acceleration changes, we'll unleash the power of Calculus to differentiate and integrate our way to the answer. Let's get moving!

Interpreting Kinematic Graphs: Displacement-Time and Velocity-Time4.2

Alright, let's decode these kinematic graphs. Think of them as the story of an object's journey, and mastering them is a massive advantage for your exams and a foundational skill for any STEM degree, from engineering to physics. First up, the Displacement-Time (s-t) graph. This graph plots displacement ($s$) against time ($t$). The single most important takeaway here is: the gradient is velocity. Remember, gradient is 'rise over run', or $\frac{\Delta s}{\Delta t}$, which is literally the definition of velocity ($v$). A steep line? You're moving fast. A flat, horizontal line? You're stationary (zero velocity), probably scrolling on your phone. A line with a negative gradient means you're moving back towards your starting point. A curve means your velocity is changing, which means you're accelerating! A displacement-time graph with three sections: a straight positive slope labelled 'Constant positive velocity', a horizontal line labelled 'Stationary (zero velocity)', and a curve getting steeper labelled 'Accelerating'. Now for the main even . This concept is fundamental in fields like vehicle dynamics and aerospace engineering—they use this to model everything from F1 cars to spacecraft trajectories. So yeah, it's pretty important! 😉

Worked Example: Multi-Stage Journey Analysis4.2

Problem A cyclist accelerates uniformly from rest to a speed of $8 \text{ m s}^{-1}$ in $10$ seconds. She then travels at this constant speed for $20$ seconds, before decelerating uniformly to rest in a further $16$ seconds. Find: (a) the acceleration in the first 10 seconds, (b) the total distance travelled, and (c) the deceleration in the final 16 seconds.

Step 1 First things first, let's sketch the v-t graph. This is a critical problem-solving tool that helps you visualize the entire journey. We have three distinct phases: acceleration (a straight line from (0,0) to (10,8)), constant velocity (a horizontal line from (10,8) to (30,8)), and deceleration (a straight line from (30,8) down to (46,0)). The total time is $10 + 20 + 16 = 46$ seconds. A sketch of the velocity-time graph, a trapezium with vertices at (0,0), (10,8), (30,8), and (46,0). The x-axis is labelled 'time (s)' and the y-axis is 'velocity (m/s)'.

Step 2 For part (a), we need to find the initial acceleration. Remember, acceleration is simply the gradient of the v-t graph. For the first 10 seconds, we're looking at the gradient of the first line segment. $$a = \frac{\text{change in velocity}}{\text{change in time}} = \frac{8 - 0}{10 - 0} = 0.8 \text{ m s}^{-2}$$

Step 3 For part (b), we need the total distance travelled. This is the total area under the v-t graph. We can split the shape into a triangle, a rectangle, and a final triangle, then sum their areas. This is often the safest method to avoid errors. $$s = \left(\frac{1}{2} \times 10 \times 8\right) + (20 \times 8) + \left(\frac{1}{2} \times 16 \times 8\right) = 40 + 160 + 64 = 264 \text{ m}$$

Step 4 Finally, for part (c), we calculate the deceleration. This is the gradient of the final line segment. We expect a negative value for acceleration, and deceleration is the positive magnitude of this value. $$a = \frac{\text{change in velocity}}{\text{change in time}} = \frac{0 - 8}{46 - 30} = \frac{-8}{16} = -0.5 \text{ m s}^{-2}$$ The negative sign confirms it's a deceleration. Therefore, the deceleration is $0.5 \text{ m s}^{-2}$.

Calculus Applications in Kinematics: Variable Acceleration4.2

Alright, let's level up from the SUVAT equations. SUVAT is awesome, but it has one major limitation: it only works for constant acceleration. In the real world, things rarely move like that. Think about a car in city traffic – it's constantly speeding up and slowing down. That's variable acceleration, and to handle it, we need the power of calculus. It's the key to describing motion that isn't perfectly uniform, which is crucial for fields like engineering, robotics, and even astrophysics. The whole concept boils down to a beautiful, logical hierarchy. Displacement ($s$), velocity ($v$), and acceleration ($a$) are all linked by differentiation and integration. Think of it like a ladder. To go down the ladder from displacement to velocity, and then from velocity to acceleration, we differentiate with respect to time ($t$). Velocity is the rate of change of displacement: $$v = \frac{ds}{dt}$$ Acceleration is the rate of change of velocity: $$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$ So, if you're given an equation for displacement like $s = t^3 + 2t$, you can find the velocity and acceleration at any instant just by differentiating. Easy points in an exam! A pyramid or ladder diagram. At the top is Displacement (s). An arrow pointing down labelled 'Differentiate (d/dt)' points to Velocity (v). Another arrow pointing down from v, also labelled 'Differentiate (d/dt)', points to Acceleration (a). Then, sh Now, what about going up the ladder? If you have acceleration and want to find velocity, or have velocity and want to find displacement, you integrate. It's the reverse process. $$v = \int a \, dt \quad \text{and} \quad s = \int v \, dt$$ The one thing you absolutely cannot forget here is the constant of integration, `+ C`. This little guy is a classic exam trap! 🚨 You'll always be given some initial conditions in the problem, like 'the particle starts from rest at the origin'. You use this information (e.g., at $t=0$, $v=0$ and $s=0$) to find the specific value of 'C' for that scenario. Master this, and you're well on your way to acing these kinematics questions.

Worked Example: Motion with Variable Acceleration4.2

Problem A particle P moves along a straight line. At time $t$ seconds, its velocity, $v$ m/s, is given by $v = 2t^2 - 16t + 30$. (a) Find the acceleration of the particle when $t=5$. (b) Find the values of $t$ for which the particle is instantaneously at rest. (c) Given that the particle is at the origin when $t=0$, find the displacement of the particle from the origin when $t=3$.

Step 1 Part (a): To get acceleration from velocity, we need to differentiate the velocity function with respect to time, $t$. Remember, $a = \frac{dv}{dt}$. $$a = \frac{d}{dt}(2t^2 - 16t + 30) = 4t - 16$$ Now, we just substitute $t=5$ into our new acceleration equation: $$a(5) = 4(5) - 16 = 20 - 16 = 4 \, \text{m/s}^2$$

Step 2 Part (b): The phrase 'instantaneously at rest' is a key trigger. It means the velocity is zero, so we set $v=0$ and solve the resulting quadratic equation for $t$. $$2t^2 - 16t + 30 = 0$$ Divide by 2 to simplify: $$t^2 - 8t + 15 = 0$$ Factorise the quadratic: $$(t-3)(t-5) = 0$$ $$t=3 \, \text{s and } t=5 \, \text{s}$$

Step 3 Part (c): To find displacement ($s$) from velocity ($v$), we must integrate the velocity function. This is where we go 'up' the ladder. Don't forget the constant of integration, `+ C`! $$s = \int v \, dt = \int (2t^2 - 16t + 30) \, dt$$ $$s = \frac{2t^3}{3} - \frac{16t^2}{2} + 30t + C$$ $$s = \frac{2}{3}t^3 - 8t^2 + 30t + C$$

Step 4 Now we find the value of C. The problem states that the particle is at the origin ($s=0$) when $t=0$. We substitute these values into our displacement equation to solve for C. $$0 = \frac{2}{3}(0)^3 - 8(0)^2 + 30(0) + C$$ $$C = 0$$ So, our specific equation for displacement is: $$s = \frac{2}{3}t^3 - 8t^2 + 30t$$

Step 5 Finally, we can answer the question. We need to find the displacement when $t=3$. We simply substitute $t=3$ into our complete displacement equation. $$s(3) = \frac{2}{3}(3)^3 - 8(3)^2 + 30(3)$$ $$s(3) = \frac{2}{3}(27) - 8(9) + 90$$ $$s(3) = 18 - 72 + 90 = 36 \, \text{m}$$

Using Constant Acceleration Formulae4.2

Alright, let's talk about the SUVAT equations. Think of them as the ultimate cheat codes for solving motion problems, but like, totally legit ones your examiner wants you to use. This only works for a specific scenario, though: when an object is moving with constant acceleration. If the acceleration is changing, these are a no-go. It’s like trying to use a Wi-Fi password for a different network – it just won't connect. So, what does SUVAT even stand for? It's an acronym for the five variables we use to describe motion: $s$ = displacement (how far you are from the start, with direction) $u$ = initial velocity (your speed at the start) $v$ = final velocity (your speed at the end) $a$ = acceleration (how quickly your velocity changes) $t$ = time The magic is that if you know any three of these, you can find the other two using a set of special formulae. It's like a group chat where if three people have seen the message, you can figure out what the other two are up to. A diagram of a car moving in a straight line. Arrows and labels clearly show the initial velocity 'u' at the start point, the final velocity 'v' at the end point, the displacement 's' between the points, and a clock icon indicating the time 't' taken Here are the three heavy-hitters you need to know: 1. $v = u + at$ (Perfect when you don't care about displacement, $s$) 2. $s = ut + \frac{1}{2}at^2$ (Your go-to when final velocity, $v$, is the missing piece of the puzzle) 3. $v^2 = u^2 + 2as$ (The MVP when time, $t$, hasn't been invited to the party) Your main skill here is playing detective: read the problem, list out your known SUVAT values, identify the unknown you need to find, and then pick the one equation that links them all together. Get this down, and you'll be flying through kinematics questions.

Worked Example: Applying SUVAT to Vertical Motion4.2

Problem You're standing on a bridge 30 m above a river. You accidentally drop your phone (don't worry, it has a waterproof case!). Assuming it's dropped from rest and that air resistance is negligible, find the speed at which the phone hits the water. Use $a = 9.8 \, \text{ms}^{-2}$ for acceleration due to gravity.

Step 1 First, let's be detectives and list our SUVAT variables. It's crucial to define a positive direction – let's say downwards is positive. $s = +30$ m (it travels 30m down) $u = 0 \, \text{ms}^{-1}$ ('dropped from rest' is the key phrase) $v = ?$ (this is what we want to find!) $a = +9.8 \, \text{ms}^{-2}$ (gravity acts downwards, our positive direction) $t = $ (we don't know it, and the question doesn't ask for it)

Step 2 Now we pick our weapon... I mean, equation. We have values for $s$, $u$, and $a$, and we need to find $v$. The variable that's missing and not needed is $t$. The only SUVAT equation that doesn't involve time is our go-to choice here. $$v^2 = u^2 + 2as$$

Step 3 Time to sub in the numbers and do the math. We're just plugging in the values we listed in Step 1. Remember that since $u=0$, the calculation is a bit simpler. $$v^2 = (0)^2 + 2(9.8)(30) \\ v^2 = 588$$

Step 4 We're almost there! The question asks for the speed ($v$), not $v^2$. So, the final move is to take the square root. Always finish by stating your answer clearly with the correct units, usually rounded to 3 significant figures. $$v = \sqrt{588} \approx 24.2 \, \text{ms}^{-1}$$

Key Points to Remember

For velocity-time graphs, the gradient is acceleration ($a = \frac{dv}{dt}$) and the area under the graph is displacement ($s = \int v \, dt$).
The five SUVAT equations are your best friends only when acceleration is constant. Always list your knowns and unknown before picking an equation!
When acceleration is variable, it's calculus time! Differentiate to go from displacement to velocity to acceleration ($s \rightarrow v \rightarrow a$), and integrate to go the other way ($a \rightarrow v \rightarrow s$).

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