Answer to Question #102633 in Electricity and Magnetism for Ali hamza

Question

A particle starts from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of -15 m/s. The particle moves in the xy plane with an x component of acceleration only, given by ax = 4.0 m/s2. (a) Determine the components of the velocity vector at any time and the total velocity vector at any time. (b) Calculate the velocity and speed of the particle at t = 5.0 s. (c) Determine the x and y coordinates of the particle at any time t and the position vector at this time.

Answer 1

Solution. According to the conditions of the problem for t=0s

$$x_0=0m$$

$$y_0=0m$$ $$v_{0x}=20 \frac{m}{s}$$

$$v_{0y}=-15 \frac{m}{s}$$ $$a_x=4 \frac{m}{s^2}$$

$$a_y=0 \frac{m}{s^2}$$

a) The particle moves with constant acceleration along the 0x axis; therefore, the component v_x can be written as

$$v_x(t)=20+4t \frac{m}{s}$$

The particle moves at a constant speed along the axis 0y; therefore, the component wu can be written as

$$v_y(t)=-15 \frac{m}{s}$$

Therefore, the total velocity vector is equal to

$$v(t)=(20+4t)i-15j$$

b)  Calculate the velocity at t = 5.0 s.

$$v(5)=(20+20)i-15j=40i-15j$$

Calculate the speed at t = 5.0 s.

$$v=\sqrt{20^2+(-15)^2}=\sqrt{625}=25 \frac{m}{s}$$

c) Integrating the velocity equations of the components by the time and using the values of the initial coordinates, we obtain

$$x(t)=20t+2t^2$$ $$y(t)=-15t$$

As result position vector at this time

$$r(t)=(20t+2t^2)i-15tj$$

Answer. a)

$$v_x(t)=20+4t \frac{m}{s}$$ $$v_y(t)=-15 \frac{m}{s}$$ $$v(t)=(20+4t)i-15j$$

b) velocity

$$v(5)=40i-15j$$

speed

$$v=25 \frac{m}{s}$$

c)

$$x(t)=20t+2t^2$$

$$y(t)=-15t$$

$$r(t)=(20t+2t^2)i-15tj$$