Question

a particle starts from rest at xi=0 and moves for 10 seconds with an acceleration of 2 cm/s^2 for the next 20 seconds the acceleration of the particle is -1 cm/s^2 what is the position of the particle at the end of the motion

Answer 1

A particle starts from rest at $x_i = 0$ and moves for 10 seconds with an acceleration of $2\,\mathrm{cm/s^2}$ for the next 20 seconds the acceleration of the particle is $-1\,\mathrm{cm/s^2}$ what is the position of the particle at the end of the motion

Solution

At the first part of the way position of the particle is giving by formula

$$x_1(t) = x_{10} + v_{10}t + \frac{a_1t^2}{2} = 0 + 0*t + \frac{a_1t^2}{2} = \frac{a_1t^2}{2}.$$

A position of the particle after 10 seconds:

$$x_1(10) = \frac{2\,\frac{\mathrm{cm}}{\mathrm{s}^2}*(10s)^2}{2} = 100\,\mathrm{cm}.$$

At the first part of the way position of the particle is giving by formula

$$x_2(t) = x_{20} + v_{20}t + \frac{a_2t^2}{2}.$$ $$x_{20} = x_1(10) = 100\,\mathrm{cm},\ v_{20} = 2\,\frac{\mathrm{cm}}{\mathrm{s}^2}*10s = 20\,\frac{\mathrm{cm}}{\mathrm{s}}.$$

The position of the particle at the end of the motion

$$x_2(20) = 100 + 20*20 + \frac{(-1)*20^2}{2} = 300\,\mathrm{cm}.$$

Answer: 300 cm.