Click to Chat

1800-2000-838

+91-120-4616500

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 = 3t2-2t3
At 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?)```
8 years ago

21 Points
```										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
```
8 years ago
Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details