Question

The position of a particle moving along the x axis is given by x = 6.0t2  1.0t3, where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

Answer 1

Given that,

x=9t  

2

−t  

3

We know that,

v=  

dt

dx

​  

 

The maximum speed is

v=18t−3t  

2

 

v maximum, at  

dt

dv

​  

=0

dt

dv

​  

=18−6t

The time will be

0=18−6t

t=3 sec

The position of the particle at 3 sec

Put the value of t in equation (I)

x=9×9−27

x=81−27

x=54 m

The position of the particle will be 54 m.

Hence, Option A is correct

Answer 2

The position of the particle when it achieves its maximum speed in the positive x-direction is 16 meter.

Given:

$x=6.0t^2-1.0t^3$

To find:

The position of the particle when it achieves its maximum speed in the positive x direction

Solution:

We can write $x=6.0t^2-1.0t^3$ as

$x=6t^2-t^3$                                          ...(I)

The speed of the particle is given by,

$v=\frac{dx}{dt}$  where v is speed.

(Note: Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object’s movement. There is difference between the two. )

Taking derivative of equation (I) with respect to t,

$\frac{dx}{dt} =12t-3t^2$

The maximum speed is

$v=12t-3t^2$

Speed is maximum means acceleration becomes zero as there is no change in speed.

v maximum, at  

$\frac{dv}{dt}=0$                                                   ...(II)

​Calculate $\frac{dv}{dt}$ ,

​$\frac{dv}{dt}=12-6t$                                         ...(III)

Calculate value of t:

  0=12−6t

⇒t=2 sec                                             ...(By II and III)

To calculate the position of the particle when it achieves its maximum speed in the positive x-direction put t=3 in (I)

$x=6t^2-t^3$

   $=6(2)^{2} -(2)^{3}$

   =(6×4) - (8)

   = 24 - 8

   =16

x=16 meter

Hence,the position of the particle when it achieves its maximum speed in the positive x direction is 16 meter.

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