#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# The position of a particle moving along the x axis depends on the time according to equation x = 3t2-2t3At what time does the particle reach its maximum positive x position? And when does it reach minimum?(I believe this requires calculus. Could someone show me?)

12 years ago

Hi Pritish,

Remember that at the point of maxima and minima, dy/dx = 0

and if d2y/dx2 is -ve, then that value of x gives maxima and if,

d2y/dx2 = +ve, then it is minima.

Given, x = 3t2 - 2t3

dx/dt = 6t - 6t2

For maxima, or minima,

dx/dt = 0

6t - 6t2 = 0

t = t2

t (t - 1) = 0

t = 0s & t = 1 sec

Let us find d2x/dt2

(d2x/dt2 )t=0  6 - 12t

=   6 - 12 x 0

=  6+ve

(d2x/dt2 )t=1  =  6 - 12 x 1

= -6

=  - ve

Hence, at t = 1 sec, particle reaches its maximum +ve  x position and at x = 0 it reaches minimum x - position