The point moves along the x-axis with constant acceleration a = –3 m / s ^ 2

The point moves along the x-axis with constant acceleration a = –3 m / s ^ 2, at the initial moment of time its speed is +3 m / s. Determine the path traveled by the point in 4 s.

The formula connecting all the basic quantities that are significant for describing rectilinear uniformly accelerated motion is as follows:

S = v start * t + (аt²) / 2, where S is the path (or distance) traveled by the body, v initial is the initial speed, t is the time the body is on the way, and is the constant acceleration with which it was gaining speed.

In our case, the acceleration is “negative”, that is, its vector (direction) is directed opposite to the vector (direction) of speed.

This can be compared to the movement of a body thrown vertically upward, against which the acceleration of gravity acts. Thus, the path formula will look like this:

S = v start * t – (аt²) / 2.

Substituting the values ​​given to us, we get –

S = 3 m / s * 4 s – ((3 m / s²) * (4 s) ²) / 2 = 12 m – 24 m = – 12 m.

Since the condition says that the point moves along the x-axis, the result obtained indicates that 4 seconds after the point starts moving, its movement will be 12 meters (modulo) in the opposite direction (with respect to the velocity vector). (Strictly speaking, in the first second of its path, the body managed to go 1.5 m forward before it stopped and “went back”, so the path covered by the body will be 13.5 meters …)