Question

answer me plzzzz guys.........

Answer 1

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

Sx = 16m

Component of initial velocity in x direction is

ux = 0ms-1

component of acceleration in x direction is

ax = 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 Sx , ux , ax 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

uy = 10ms-1

acceleration of the particle y direction is

ay = 2ms-2

time taken by the particle

t = 2s

Displacement of the particle in y direction

Sy = ?

Again, applying the equation of motion in


Putting the values of uy,t,ay to find Sy


Sy = 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

ux = 0 m/s

component of acceleration in x direction is

ax = 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

vx = ?

so applying the equation of motion

vx = ux + axt

after putting the values, we get

vx = 0 + 8ms-2×2s

vx = 0 + 16 m/s = 16 m/s

so the x component of final velocity of the particle is

vx = 16 m/s

Component of initial velocity of particle in Y direction

uy = 10 m/s

component of acceleration in y direction is

ax = 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

vy = ?

applying the equation of motion

vy = uy + ayt

putting the values we get

vy = 10ms-1+ 2ms-2×2s

vy = 10ms-1+ 4ms-1

vy = 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 vx and vy 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