here the velocity and acceleration of particle are given in vector form, and are unit vectors along X axis and Y axis respectively

i.e they are vectors with magnitude unity and are representing X direction and Y direction, each vector in x-y plane can be resolved into two components one along x direction and other along y direction, the vector quantity is the linear combination of both the components

initial velocity of the particle is

i.e its magnitude is 10.0 m/s in y direction

since coefficient of is zero so magnitude in x direction is zero

the acceleration of the particle is

i.e. the magnitude in x direction is 8.0 m/s and magnitude along y direction is 2.0 m/s

(a) since the particle started from origin so its displacement in x direction will give its new x co-ordinate and displacement in y direction will give its new y co-ordinate, so the displacement of particle in x direction is

S_x = 16m

Component of initial velocity in x direction is

u_x = 0ms^-1

component of acceleration in x direction is

a_x = 8ms^-2

we have to find time taken by particle

t=?

we will use the equation of motion containing these four variables

i.e.

where S denotes displacement of particle, u denotes initial velocity of particle ,a denotes acceleration of particle and t denotes the time taken to reach the displacement

since we have to find time taken when displacement in x direction is given so modifying the equation

And putting the values of S_x , u_x , a_x in above equation find the value of t we get,

i.e. when x co-ordinate of particle is 16m when time t = 2s

now we have to find the Y Co-ordinate of particle, that’s the displacement of particle in y direction when particle moved 16 m in x direction at the end of 2s

Now initial velocity of particle in x direction is

u_y = 10ms^-1

acceleration of the particle y direction is

a_y = 2ms^-2

time taken by the particle

t = 2s

Displacement of the particle in y direction

S_y = ?

Again, applying the equation of motion in

Putting the values of u_y,t,a_y to find S_y

S_y = 20m + 4m = 24m

So the y co-ordinate of the particle is 24m

(b) Speed is the magnitude of velocity of particle so we will first find component of velocity in x and y direction using the equation of motion

v = u + at

Component of initial velocity of particle in x direction

u_x = 0 m/s

component of acceleration in x direction is

a_x = 8ms^-2

time after which we have to find velocity

t = 2 s

we have to find x component of final velocity of the particle

v_x = ?

so applying the equation of motion

v �_x = u_x + a_xt

after putting the values, we get

v �_x = 0 + 8ms^-2×2s

v_x = 0 + 16 m/s = 16 m/s

so the x component of final velocity of the particle is

v_x = 16 m/s

Component of initial velocity of particle in Y direction

u_y = 10 m/s

component of acceleration in y direction is

a_x = 2ms^-2

time after which we have to find velocity

t = 2 s

we have to find y component of final velocity of the particle

v_y = ?

applying the equation of motion

v �_y = u_y + a_yt

putting the values we get

v_y = 10ms^-1+ 2ms^-2×2s

v_y = 10ms^-1+ 4ms^-1

v_y = 14ms^-1

so magnitude of velocity of particle in y direction is 14 m/s

now the speed of the particle or the magnitude of velocity of the particle is

Putting the values of v_x and v_y we get

So the speed of the particle is 21.26 m/s

The position and speed of particle at t = 2 sec are shown in the figure