Question

The position X of a particle moving along a straight line path varies with time according to the relation X =6t square -5t

Answer 1

Given, x

2

=2t

2

+6t+1 ...(i)

Differentiating Eq. (i) w.r.t. t, we get

2x

dt

dx

=4t+6

2xv=4t+6 (

∵v=

dt

dx

)

xv=2t+3 ...(ii)

Now, again differentiating Eq.(ii) w.r.t. t, we get

x

dt

dv

+v

dt

dx

=2

x.a+v.v=2 (

∵a=

dt

dv

and v=

dt

dx

)

xa+v

2

=2 ...(iii)

Here, v

2

=

x

2

4t

2

+12t+9

v

2

=

x

2

2(2t

2

+6t+1)+7

v

2

=

x

2

2x

2

+7

...(iv)

Put the value of v

2

in Eq. (iii) from Eq. (iv), we get

xa+

x

2

2x

2

+7

=2

x

3

a+2x

2

+7=2x

2

x

3

a+7=0

x

3

a=−7

a=

x

3

−7

Hence, the acceleration varies with x

−3

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