Such motion arises, according to Newton’s law, when the resultant force acting on a body, speeding it up or slowing it down, is constant.

Such conditions arise rather frequently, even though only approximately: a car moving with its motor cut off slows down under the action of the more or less constant force of friction: a weighty object falls from a height under the action of the constant force of gravity.

Knowing the magnitude of the resultant force, and also the mass of a body, we can find the magnitude of the acceleration according to the formula \(a = F/m\). Since \[a= \frac{v - v_{0}}{t}\] where \(t\) is the time of the motion, \(v\) is the final speed, and is the initial speed, with the aid of this formula it is possible to answer a series of questions of, say, the following type: How long will it take a train to come to a halt if the decelerating force, the mass of the train and the initial speed are known? Or how much speed will a car gather if the power of the motor, the resistance, the mass of the car and the duration of acceleration are known?

We are often interested in knowing the distance covered by a body in a uniformly accelerated motion. If the motion is uniform, the distance covered is found by multiplying the speed of the motion by its time. If the motion is uniformly accelerated, the calculation of the distance covered is carried out as though the body were moving uniformly for the same time \(t\) with the speed equal to half the sum of the initial and final speeds: \[s = \frac{1}{2} (v + v_{0}) \, t\] Thus, for uniformly accelerated (or decelerated) motion, the distance covered by a body is equal to the product of half the sum of the initial and final speeds by the time of the motion. The same distance would be covered during the same time in a uniform motion with speed \((v_{0}+v)/2\). In this sense, one can say that \((v_{0}+v)/2\) is the average speed of the uniformly accelerated motion.

It is helpful to compose a formula which would show the dependence of the distance covered on the acceleration. Substituting \(v = v_{0} + at\) in the last formula, we find: \[s = v_{0}t + \frac{1}{2} a t^{2}\] or, if the motion occurs without any initial speed, \[s = \frac{1}{2} a t^{2}\] If a body travels 5 m in one second, then in two seconds it will travel \((4 \times 5) \, \mathrm{m}\), in three seconds \((9 \times 5) \, \mathrm{m}\), etc. The distance travelled grows in proportion to the square of the time.

A heavy body falls from a height in accordance with this law. The acceleration of free fall is equal to \(g\), and our formula acquires the following form: \[s = \frac{981}{2} t^{2}\] if \(t\) is expressed in seconds and \(g\) in centimetres per second per second.

If a body could fall without hindrance for some 100 s, it would cover an enormous distance from the beginning of its fall—about 50 km. Moreover, only a mere 0.5 km would be covered in the first 10 s—this is what accelerated motion means.

But what speed will a body develop in falling from a given height? To answer this question we shall need formulas relating the covered distance to the acceleration and the speed. Substituting the time of the motion \(t = (v - v_{0})/a\) in \(s = (1/2)(v_{0}+v) t\), we obtain: \[s = \frac{1}{2a}(v^{2} - v_{0}^{2}) \label{dist-time-acc}\] or, if the initial speed is equal to zero, \[s = \frac{v^{2}}{2a} \quad v =\sqrt{2as}\] Ten metres is the height of a small two- or three-storey house. Why is it dangerous to jump to the ground from the roof of such a house? A simple calculation shows that the speed of such a free fall would reach the value \(v = \sqrt{2 \times 9.8 \times 10} \, \mathrm{m/s} = 14\, \mathrm{m/s} \approx 50 \mathrm{km/h}\), and this is, after all, the speed of a car within city limits. Air resistance will not reduce this speed much.

The formulas we have singled out are employed for the most varied computations. Let us apply them in order to see how motions take place on the Moon.

In H. G. Wells’ novel The First Men in the Moon we read about the surprises experienced by travellers in their fantastic trips. On the Moon, the acceleration of gravity is approximately six times less than terrestrial. If a falling body on the Earth covers 5 m in the first second, it will “float” down only 80 cm in all on the Moon (the acceleration there is about 1.6 \(\mathrm{m/s^2}\)).

The formulas we have written out permit us to rapidly calculate the lunar “miracles”.

A jump from a height of \(h \,\mathrm{m}\) takes \(t = \sqrt{2h/g}\,s\). Since lunar acceleration is six times less than terrestrial, the jump will require \(\sqrt{6} \approx 2.45\) times more time on the Moon. By how many times will the final speed of the jump be: decreased \((v = \sqrt{2gh})\)?

One can jump safely from the roof of a three-story house on the Moon. The height of a jump with the same initial speed will be increased by a factor of six \((h = v^{2}/2g)\). A child will be able to jump higher than the record set on the Earth.