Chapter 2 ยท Introduction

Chapter 2 โ€” Motion in a Straight Line

Where is it, how fast is it going, and how quickly is that changing? This chapter turns those three questions into position, velocity and acceleration โ€” and reads the answers off graphs.

A driver is doing $50\ \text{km/h}$ past a school when a child steps out between parked cars $30\ \text{m}$ ahead. It takes the driver about one second to react and hit the brake, and the car then slows at about $7\ \text{m/s}^2$ on a dry road. It stops after $27.7\ \text{m}$ โ€” just short of the child. Now replay the same moment at $60\ \text{km/h}$, only $10\ \text{km/h}$ faster. The car covers $16.7\ \text{m}$ before the brakes even come on, and it is still moving at about $34\ \text{km/h}$ when it reaches the child. Nothing about the driver or the brakes changed; only the speed did.

Why does a small change in speed make such a large change in stopping distance? The answer lies in a few simple ideas that Galileo pieced together by rolling balls down gentle ramps four centuries ago: the position of a moving body, its velocity (how fast the position changes), and its acceleration (how fast the velocity changes). In this chapter you will define each of them precisely โ€” using the language of rates, $\dfrac{dx}{dt}$, for the first time โ€” learn to read them from xโ€“t, vโ€“t and aโ€“t graphs, discover that the area under a velocityโ€“time graph is a distance, and derive the equations that predict braking distances, chases and the flight of a ball thrown straight up.

Position & frameโ†’Displacementโ†’Velocity $\tfrac{dx}{dt}$โ†’Acceleration $\tfrac{dv}{dt}$โ†’xโ€“t, vโ€“t, aโ€“t graphsโ†’Area = displacementโ†’Kinematic equationsโ†’Free fall
๐ŸŽฏ By the end of this chapter you canโ€ฆ
  • choose a frame of reference and describe straight-line motion by a signed position coordinate; distinguish path length from displacement;
  • define instantaneous velocity as $v=\dfrac{dx}{dt}$, find it as the slope of an xโ€“t graph, and compute it from a formula for $x(t)$;
  • define acceleration as $a=\dfrac{dv}{dt}$, decide from the signs of $v$ and $a$ whether a body is speeding up or slowing down, and sketch matching xโ€“t, vโ€“t and aโ€“t graphs;
  • find displacement and distance as areas under a vโ€“t graph, and change in velocity as the area under an aโ€“t graph;
  • derive the kinematic equations for constant acceleration (by graphs and by calculus) and apply them to braking, chases, free fall and vertical throws;
  • solve Board, JEE Main, JEE Advanced and NEET-style problems on one-dimensional motion.
Teaser โ€” Stopping in Time

A child is $30\ \text{m}$ ahead when the driver notices. The car keeps its speed during the reaction time, then brakes uniformly. Set the speed, the reaction time and the road, and press Play. The lower graph is the car's velocity against time: the shaded area under it is exactly the distance travelled โ€” you will see why in Section 2.2.

60 km/h1.0 s
reaction distanceโ€”
braking distanceโ€”
stopping distanceโ€”
speed on reaching the childโ€”

Sections in this chapter

โœ… Before you start
From Class 9 Science (Motion) recall distance, speed and the meaning of a distanceโ€“time graph. From mathematics you need the slope of a straight line, the areas of a rectangle, triangle and trapezium, and solving a quadratic equation. From Chapter 1 you need SI units and conversions such as $1\ \text{km/h}=\tfrac{1}{3.6}\ \text{m/s}$ (Section 1.1), significant figures (Section 1.2), and dimensional checks of equations such as $x=ut+\tfrac12at^2$ (Section 1.3). The calculus used here โ€” rates of change and areas under graphs โ€” is introduced as it is needed.