__Question 4.21__:

A particle starts from the origin at t = 0 s with a velocity of 10.0

**j**m/s and moves in the x-y plane with a constant acceleration of (8.0

**i**+ 2.0

**j**) m s

^{-2}.

(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time ?

__Solution__:

Velocity of the particle,

**v**= 10.0

**j**m/s

Acceleration of the particle

**a**= (8.0

**i**+ 2.0

**j**)

Also,

But,

**a**= d

**v**/ dt = 8.0

**i**+ 2.0

**j**

d

**v**= (8.0

**i**+ 2.0

**j**) dt

Integrating both sides:

**v**(t) = 8.0t

**i**+ 2.0t

**j**+

**u**

where,

**u**= Velocity vector of the particle at

*t*= 0

**v**= Velocity vector of the particle at time

*t*

*But,*

**v**

*=*d

**r**/ dt

d

**r**=

**v**dt = (8.0t

**i**+ 2.0t

**j**+

**u**) dt

Integrating the equations with the conditions: at

*t*= 0;

*= 0 and at*

**r***t = t*;

*r = r*

**r**=**u**t + (1/2) X 8.0 t

^{2}

**i**+ (1/2) X 2.0 t

^{2}

**j**

=

**u**t + 4.0 t

^{2}

**i**+ t

^{2}

**j**

= (10.0

**j**)t + 4.0t

^{2}

**i**+ t

^{2}

**j**

x

**i**+ y

**j**= 4.0t

^{2}

**i**+ (10t + t

^{2})

**j**

Since the motion of the particle is confined to the

*x-y*plane, on equating the coefficients of

**i**and

**j**, we get:

x = 4t

^{2}

t = (x/4)

^{1/2}

And y = 10t + t

^{2}

**(a)**When

*x*= 16 m:

t = (16/4)

^{1/2}= 2 s

∴

*y*= 10 × 2 + (2)

^{2}= 24 m

**(b)**Velocity of the particle is given by:

**v**(t) = 8.0t

**i**+ 2.0t

**j**+

**u**

**v**(t) = 8.0t

**i**+ 2.0t

**j**+ 10

**j**

**v**(t) = 8.0t

**i**+ (10 + 2.0t)

**j**

At t = 2 s

**v**(2) = 8 X 2

**i**+ (10 + 2 X 2)

**j**

**v**(2) = 16

**i**+ 14

**j**

**|**

**v**(2)

**|**= ( (16)

^{2}+ (14)

^{2 })

^{1/2}= 21.26 m/s

^{}

Speed at t = 2 s = 21.26 m/s

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