Question

A material point begins to move along the straight line with constant acceleration from its resting state. During the third second of its motion the speed of material point has increased by 2 m/s. What distance did the material point travel during the fourth second? Give your answer in meters accurate to the whole.

Answer 1

The speed under acceleration can be found from the following equation:

$$v_f(t)=v_0+at.$$

Thus, before the third second the speed was

$$v_f(2)=v_0+at=0+2a=2a.$$

By the end of the third second, the speed is

$$v_f(3)=v_f(2)+a\cdot1=3a.$$

During the third second, the speed increased by 2 m/s:

$$v_f(3)-v_f(2)=3a-2a=2,\\a=2.$$

Now, we have all we need to find the distance:

$$d(t)=v_0t+at^2/2,\\ d(4)=0\cdot4+2\cdot4^2/2=16\text{ m},\\ d(3)=0\cdot3+2\cdot3^2/2=9\text{ m},\\ \Delta d_4=d(4)-d(3)=16-9=7\text{ m}.$$