Uniformly accelerated motion
A motion is called uniformly accelerated when the acceleration stays constant. Constant means unchanged in magnitude and unchanged in direction. For such a motion, the position and the velocity at any time can be written with three short equations. These are the equations of motion. They are used more than any other equations in this book. Every one of them rests on one condition: the acceleration must be constant throughout the interval.
Five symbols appear in the equations. \(u\) is the velocity at the start of the interval. \(v\) is the velocity at the end. \(a\) is the constant acceleration. \(t\) is the duration of the interval. \(s\) is the displacement during the interval, measured from the position at the start. All five carry signs along the chosen positive direction.
The three equations of motion describe an interval during which the acceleration is constant in magnitude and in direction. If the acceleration changes at any moment inside the interval, none of the three equations applies to that interval as a whole.
The first equation
Acceleration is the rate of change of velocity. If the acceleration is a constant \(a\), the velocity changes by \(a\) in every second. In a time \(t\) the velocity changes by \(a t\). The derivation below writes this idea with calculus, so that the role of the constant acceleration is clear.
\[ a = \dfrac{dv}{dt} \]
Why. this is the definition of acceleration. For uniformly accelerated motion, \(a\) is a constant number.
\[ \int_{u}^{v} dv = \int_{0}^{t} a\,dt \]
Why. separate the variables and integrate both sides over the same interval. The velocity runs from \(u\) to \(v\) while the time runs from \(0\) to \(t\).
\[ v - u = a\,t \]
Why. a constant comes out of the integral, and the integral of \(dt\) from \(0\) to \(t\) is \(t\).
\[ v = u + at \]
Read it. the velocity after time \(t\) is the initial velocity plus the change \(at\). On a v-t graph this is a straight line of slope \(a\) starting at height \(u\).
(a) A car starts from rest and accelerates at \(2\,\text{m/s}^2\) for \(5\,\text{s}\). Find its final velocity. (b) A train moving at \(20\,\text{m/s}\) brakes with a constant deceleration of \(1\,\text{m/s}^2\). Find its velocity after \(8\,\text{s}\).
(a) \(v = u + at = 0 + 2 \times 5 = 10\,\text{m/s}\).
(b) Take the direction of motion as positive. Then \(u = +20\,\text{m/s}\) and \(a = -1\,\text{m/s}^2\), because the acceleration opposes the motion. \(v = 20 - 1 \times 8 = 12\,\text{m/s}\). The train has slowed from \(20\) to \(12\,\text{m/s}\) but is still moving forward.
The second equation, seen as an area
The area under a velocity-time graph equals the displacement. For constant acceleration the graph is a straight line from height \(u\) at \(t = 0\) to height \(v = u + at\) at time \(t\). The area under this line is a trapezium. The trapezium can be cut into two simple shapes. Think of the answer before you use the chips in Fig 6.1.
The rectangle has width \(t\) and height \(u\). Its area is \(ut\). This is the displacement the body would have if it kept its starting velocity. The triangle has base \(t\) and height \(v - u = at\). Its area is \(\tfrac12 \times t \times at = \tfrac12 at^2\). This is the extra displacement produced by the acceleration. The total displacement is the sum.
The two terms of the second equation answer two separate questions. How far would the body go at its starting speed? That is \(ut\). How much further does the acceleration carry it? That is \(\tfrac12 at^2\). When \(u = 0\) only the second term remains, and the displacement grows as the square of the time.
The third equation
Many problems give the velocities and the displacement but say nothing about the time. The third equation is made for such problems. It is obtained from the first two equations by removing \(t\). One step of the derivation below is left for you.
\[ s = \left(\dfrac{u + v}{2}\right) t \]
Why. for constant acceleration, the average velocity is the mean of the initial and final velocities, and displacement is average velocity times time.
\[ t = \dfrac{v - u}{a} \]
Why. rearrange the first equation \(v = u + at\) to make \(t\) the subject.
?
Your turn. substitute the expression for \(t\) from step 2 into step 1. Which line is it?
\[ 2as = v^2 - u^2 \]
Why. multiply out: \((u + v)(v - u) = v^2 - u^2\), then multiply both sides by \(2a\).
\[ v^2 = u^2 + 2as \]
Read it. the time has disappeared. Only the velocities, the acceleration and the displacement remain.
A car moving at \(20\,\text{m/s}\) brakes with a constant deceleration of \(5\,\text{m/s}^2\). How far does the car travel before it stops?
Direction of motion positive. \(u = +20\,\text{m/s}\), \(v = 0\), \(a = -5\,\text{m/s}^2\). The time is not given and not wanted.
\(v^2 = u^2 + 2as\), because \(t\) is absent.
\(0 = 400 + 2(-5)s\), so \(10 s = 400\) and \(s = 40\,\text{m}\).
The car travels \(40\,\text{m}\) before stopping. Check: doubling the speed to \(40\,\text{m/s}\) would give \(160\,\text{m}\), four times as far, because \(s\) grows as \(u^2\).
Which equation to use
Each equation leaves out one of the five quantities. Choose the equation that does not contain the quantity you neither know nor want.
A car passes a signal at \(2\,\text{m/s}\) and accelerates at a steady \(1\,\text{m/s}^2\). After \(1\,\text{s}\) it has moved \(2.5\,\text{m}\); after \(2\,\text{s}\), \(6\,\text{m}\); after \(4\,\text{s}\), \(16\,\text{m}\). The distances covered in successive seconds grow, because the velocity grows.
A particle moves along a line with initial velocity \(u = +6\,\text{m/s}\) and a constant acceleration \(a = -2\,\text{m/s}^2\). What is its velocity after \(5\,\text{s}\)?
You know the initial velocity, the final velocity and the acceleration of a car, but not the time. Which equation gives the distance travelled directly?
Two cars start from rest at the same moment. Car A accelerates at \(2\,\text{m/s}^2\) and car B at \(4\,\text{m/s}^2\). After the same time \(t\), how do their displacements compare?
Where the equations stop working
All three equations were derived from one assumption. The acceleration stays constant for the whole interval. If the acceleration changes at any moment inside the interval, the equations do not apply to that interval. A body in free fall with air resistance ignored has constant acceleration \(g\). A car whose driver presses the brake harder and harder does not. A ball bouncing on the floor has a very different acceleration during the bounce than during the flight. In such cases, split the motion into intervals over which the acceleration is constant, and apply the equations to each interval separately.
Every time you write \(v = u + at\), you are claiming that the acceleration was constant during the whole time \(t\). Check the claim before you use the equation. A common error is to apply one equation across a bounce, a collision, or a change of gear.
Anti-lock braking systems release and re-apply the brakes many times a second, so that the tyre keeps rolling instead of sliding. A rolling tyre grips the road better than a sliding tyre, so the deceleration \(a\) is larger. By the third equation, the stopping distance is \(s = u^2 / 2a\). A larger \(a\) gives a shorter stopping distance at the same speed. The pulsing that a driver feels in the pedal is the system keeping \(a\) as large as the road allows.
- Using the equations of motion when the acceleration is not constant.Split the motion into intervals of constant acceleration. Apply the equations to one interval at a time.
- Forgetting the sign of \(a\) when a body slows down.Choose a positive direction. If the acceleration opposes the velocity, \(a\) is negative.
- Confusing \(s\) with the position.\(s\) is the displacement during the interval, measured from where the body was at the start of the interval.