Two toy carts are released together on parallel tracks. Both carts reach the \(40\,\text{m}\) mark after exactly \(5\,\text{s}\). Their average velocities are therefore identical, \(8\,\text{m/s}\) each. Yet the two journeys look completely different. One cart moves at a steady pace from the start. The other starts slowly, then rushes past the finish.
Look at the arrows first. The arrow of the upper cart has the same length in every frame. Its velocity is not changing. The arrow of the lower cart starts at zero and grows. Its velocity is changing. The rate at which velocity changes with time is called acceleration. The upper cart has zero acceleration. The lower cart has a constant, non-zero acceleration.
Now look at the marks. The marks of the upper cart are equally spaced, because the cart covers equal distances in equal times. The marks of the lower cart are spaced more and more widely, because the cart covers more distance in each successive second. This widening spacing is the visible sign of acceleration.
The average acceleration of a body over a time interval is the change in the velocity of the body divided by the duration of the interval. Acceleration is a vector. On a line, its sign is the sign of the change in velocity.
(a) The velocity of a car changes from \(+20\,\text{m/s}\) to \(+30\,\text{m/s}\) in \(5\,\text{s}\). (b) A train moving at \(+30\,\text{m/s}\) slows to \(+10\,\text{m/s}\) in \(10\,\text{s}\). Find the average acceleration in each case.
(a) \(\bar{a} = (30 - 20)/5 = +2\,\text{m/s}^2\). The velocity grows by \(2\,\text{m/s}\) in every second.
(b) \(\bar{a} = (10 - 30)/10 = -2\,\text{m/s}^2\). The velocity falls by \(2\,\text{m/s}\) in every second. The negative sign says that the change in velocity points in the negative direction, opposite to the motion of the train.
Acceleration measures the rate of change of which quantity?
Instantaneous acceleration
Instantaneous acceleration is the acceleration at one instant. It is obtained from the average acceleration in the same way that instantaneous velocity was obtained from average velocity. Shrink the interval towards zero and take the limit. The result is the derivative of velocity with respect to time.
\[ a = \lim_{\Delta t \to 0} \dfrac{\Delta v}{\Delta t} = \dfrac{dv}{dt} \]
Since the velocity is itself the derivative of position, the acceleration is the derivative of the derivative of position. This is called the second derivative and is written \(d^2x/dt^2\).
Use the probe above. Between \(t = 0\) and \(t = 4\,\text{s}\) the graph rises with a constant slope, and the acceleration reading is a constant positive number. Between \(4\,\text{s}\) and \(6\,\text{s}\) the graph is flat, and the acceleration is zero although the velocity is \(8\,\text{m/s}\). After \(6\,\text{s}\) the graph falls, and the acceleration is negative. Height, slope and area are three different readings from one graph.
Find the acceleration for (a) \(x = t^3 + 2t\) and (b) \(x = 5 - 2t^2\). Positions in metres, times in seconds.
(a) \(v = dx/dt = 3t^2 + 2\). Then \(a = dv/dt = 6t\). The acceleration is not constant; it grows with time.
(b) \(v = dx/dt = -4t\). Then \(a = dv/dt = -4\,\text{m/s}^2\). The acceleration is constant and negative. The velocity becomes more and more negative as time passes.
A ball thrown straight up has zero velocity at the highest point of its flight. Is its acceleration zero there?
Does a negative acceleration mean slowing down?
Most students answer this question too quickly. Decide your answer before you look at the figure. Then use the chips to test all four cases.
A car on a straight road has a negative acceleration. Which statement is always true?
A body speeds up when its velocity and its acceleration have the same sign. A body slows down when its velocity and its acceleration have opposite signs. The sign of the acceleration alone does not decide. When the speed of a body is decreasing, we say that the body is retarding.
A particle starts with \(u = -4\,\text{m/s}\) and has a constant acceleration \(a = -2\,\text{m/s}^2\). Find its velocity at \(t = 1\,\text{s}\), \(2\,\text{s}\) and \(4\,\text{s}\). Is the particle speeding up or slowing down?
Positive direction to the right. \(u = -4\,\text{m/s}\), \(a = -2\,\text{m/s}^2\). Both are negative, so both point to the left.
\(v = u + at = -4 - 2t\).
At \(t = 1\,\text{s}\), \(v = -6\,\text{m/s}\). At \(t = 2\,\text{s}\), \(v = -8\,\text{m/s}\). At \(t = 4\,\text{s}\), \(v = -12\,\text{m/s}\).
The speeds are \(6\), \(8\) and \(12\,\text{m/s}\). The speed is rising, so the particle is speeding up, moving ever faster to the left. Velocity and acceleration have the same sign.
A lift is moving downward, and its acceleration also points downward. What is happening to the lift?
A car is reversing out of a parking place, and the speedometer reading is increasing. Taking forward as positive, what are the signs of \(v\) and \(a\)?
A Metro train brakes with a deceleration of about \(1\,\text{m/s}^2\). The train is moving forward with a backward acceleration. Velocity and acceleration have opposite signs, so the train slows down. The bodies of standing passengers tend to keep the earlier velocity while the floor slows down. So the passengers are thrown forward, and they should hold the handrail.
- Believing that zero velocity means zero acceleration.Ask whether the velocity is changing at that instant. At the top of a throw the velocity is zero but is changing, so the acceleration is not zero.
- Believing that negative acceleration means slowing down.Compare the signs of \(v\) and \(a\). Same sign, speeding up. Opposite signs, slowing down.
- Reading the height of a v-t graph as the acceleration.On a v-t graph the height is the velocity. The slope is the acceleration. The area is the displacement.